From b94604b9df885dabfedb48ace0df1246f58d362d Mon Sep 17 00:00:00 2001 From: pluveto Date: Sun, 15 Jan 2023 02:06:00 +0000 Subject: [PATCH 1/2] Reinit --- .vscode/settings.json | 3 + README.md | 10 +- src/solution/mod.rs | 237 ---------------- src/solution/s0001_two_sum.rs | 53 ---- src/solution/s0002_add_two_numbers.rs | 98 ------- ..._substring_without_repeating_characters.rs | 61 ----- .../s0004_median_of_two_sorted_arrays.rs | 62 ----- .../s0005_longest_palindromic_substring.rs | 82 ------ src/solution/s0006_zigzag_conversion.rs | 95 ------- src/solution/s0007_reverse_integer.rs | 73 ----- src/solution/s0008_string_to_integer_atoi.rs | 131 --------- src/solution/s0009_palindrome_number.rs | 89 ------ .../s0010_regular_expression_matching.rs | 92 ------- .../s0011_container_with_most_water.rs | 81 ------ src/solution/s0012_integer_to_roman.rs | 114 -------- src/solution/s0013_roman_to_integer.rs | 115 -------- src/solution/s0014_longest_common_prefix.rs | 90 ------ src/solution/s0015_3sum.rs | 143 ---------- src/solution/s0016_3sum_closest.rs | 76 ------ ...7_letter_combinations_of_a_phone_number.rs | 81 ------ src/solution/s0018_4sum.rs | 99 ------- .../s0019_remove_nth_node_from_end_of_list.rs | 71 ----- src/solution/s0020_valid_parentheses.rs | 94 ------- src/solution/s0021_merge_two_sorted_lists.rs | 75 ----- src/solution/s0022_generate_parentheses.rs | 72 ----- src/solution/s0023_merge_k_sorted_lists.rs | 95 ------- src/solution/s0024_swap_nodes_in_pairs.rs | 73 ----- .../s0025_reverse_nodes_in_k_group.rs | 106 ------- ...026_remove_duplicates_from_sorted_array.rs | 87 ------ src/solution/s0027_remove_element.rs | 97 ------- src/solution/s0028_implement_strstr.rs | 55 ---- src/solution/s0029_divide_two_integers.rs | 52 ---- ...bstring_with_concatenation_of_all_words.rs | 175 ------------ src/solution/s0031_next_permutation.rs | 89 ------ .../s0032_longest_valid_parentheses.rs | 95 ------- .../s0033_search_in_rotated_sorted_array.rs | 105 ------- ...ast_position_of_element_in_sorted_array.rs | 45 --- src/solution/s0035_search_insert_position.rs | 59 ---- src/solution/s0036_valid_sudoku.rs | 172 ------------ src/solution/s0037_sudoku_solver.rs | 51 ---- src/solution/s0038_count_and_say.rs | 81 ------ src/solution/s0039_combination_sum.rs | 100 ------- src/solution/s0040_combination_sum_ii.rs | 107 -------- src/solution/s0041_first_missing_positive.rs | 91 ------ src/solution/s0042_trapping_rain_water.rs | 38 --- src/solution/s0043_multiply_strings.rs | 67 ----- src/solution/s0044_wildcard_matching.rs | 91 ------ src/solution/s0045_jump_game_ii.rs | 45 --- src/solution/s0046_permutations.rs | 70 ----- src/solution/s0047_permutations_ii.rs | 179 ------------ src/solution/s0048_rotate_image.rs | 122 --------- src/solution/s0049_group_anagrams.rs | 67 ----- src/solution/s0050_powx_n.rs | 64 ----- src/solution/s0051_n_queens.rs | 109 -------- src/solution/s0052_n_queens_ii.rs | 99 ------- src/solution/s0053_maximum_subarray.rs | 57 ---- src/solution/s0054_spiral_matrix.rs | 101 ------- src/solution/s0055_jump_game.rs | 66 ----- src/solution/s0056_merge_intervals.rs | 85 ------ src/solution/s0057_insert_interval.rs | 126 --------- src/solution/s0058_length_of_last_word.rs | 55 ---- src/solution/s0059_spiral_matrix_ii.rs | 87 ------ src/solution/s0060_permutation_sequence.rs | 76 ------ src/solution/s0061_rotate_list.rs | 50 ---- src/solution/s0062_unique_paths.rs | 80 ------ src/solution/s0063_unique_paths_ii.rs | 111 -------- src/solution/s0064_minimum_path_sum.rs | 69 ----- src/solution/s0065_valid_number.rs | 60 ---- src/solution/s0066_plus_one.rs | 72 ----- src/solution/s0067_add_binary.rs | 79 ------ src/solution/s0068_text_justification.rs | 200 -------------- src/solution/s0069_sqrtx.rs | 72 ----- src/solution/s0070_climbing_stairs.rs | 71 ----- src/solution/s0071_simplify_path.rs | 112 -------- src/solution/s0072_edit_distance.rs | 60 ---- src/solution/s0073_set_matrix_zeroes.rs | 68 ----- src/solution/s0074_search_a_2d_matrix.rs | 92 ------- src/solution/s0075_sort_colors.rs | 84 ------ .../s0076_minimum_window_substring.rs | 56 ---- src/solution/s0077_combinations.rs | 77 ------ src/solution/s0078_subsets.rs | 67 ----- src/solution/s0079_word_search.rs | 133 --------- ..._remove_duplicates_from_sorted_array_ii.rs | 68 ----- ...s0081_search_in_rotated_sorted_array_ii.rs | 52 ---- ...2_remove_duplicates_from_sorted_list_ii.rs | 59 ---- ...0083_remove_duplicates_from_sorted_list.rs | 59 ---- .../s0084_largest_rectangle_in_histogram.rs | 91 ------ src/solution/s0085_maximal_rectangle.rs | 43 --- src/solution/s0086_partition_list.rs | 65 ----- src/solution/s0087_scramble_string.rs | 85 ------ src/solution/s0088_merge_sorted_array.rs | 72 ----- src/solution/s0089_gray_code.rs | 82 ------ src/solution/s0090_subsets_ii.rs | 95 ------- src/solution/s0091_decode_ways.rs | 52 ---- src/solution/s0092_reverse_linked_list_ii.rs | 54 ---- src/solution/s0093_restore_ip_addresses.rs | 35 --- .../s0094_binary_tree_inorder_traversal.rs | 64 ----- .../s0095_unique_binary_search_trees_ii.rs | 89 ------ .../s0096_unique_binary_search_trees.rs | 43 --- src/solution/s0097_interleaving_string.rs | 115 -------- .../s0098_validate_binary_search_tree.rs | 97 ------- .../s0099_recover_binary_search_tree.rs | 162 ----------- src/solution/s0100_same_tree.rs | 76 ------ src/solution/s0101_symmetric_tree.rs | 87 ------ ...s0102_binary_tree_level_order_traversal.rs | 81 ------ ...inary_tree_zigzag_level_order_traversal.rs | 87 ------ .../s0104_maximum_depth_of_binary_tree.rs | 61 ----- ...ree_from_preorder_and_inorder_traversal.rs | 71 ----- ...ee_from_inorder_and_postorder_traversal.rs | 78 ------ ...07_binary_tree_level_order_traversal_ii.rs | 82 ------ ...vert_sorted_array_to_binary_search_tree.rs | 64 ----- ...nvert_sorted_list_to_binary_search_tree.rs | 70 ----- src/solution/s0110_balanced_binary_tree.rs | 97 ------- .../s0111_minimum_depth_of_binary_tree.rs | 65 ----- src/solution/s0112_path_sum.rs | 80 ------ src/solution/s0113_path_sum_ii.rs | 86 ------ ...0114_flatten_binary_tree_to_linked_list.rs | 90 ------ src/solution/s0115_distinct_subsequences.rs | 122 --------- src/solution/s0118_pascals_triangle.rs | 70 ----- src/solution/s0119_pascals_triangle_ii.rs | 64 ----- src/solution/s0120_triangle.rs | 67 ----- .../s0121_best_time_to_buy_and_sell_stock.rs | 62 ----- ...0122_best_time_to_buy_and_sell_stock_ii.rs | 67 ----- ...123_best_time_to_buy_and_sell_stock_iii.rs | 106 ------- .../s0124_binary_tree_maximum_path_sum.rs | 100 ------- src/solution/s0125_valid_palindrome.rs | 72 ----- src/solution/s0126_word_ladder_ii.rs | 258 ------------------ src/solution/s0127_word_ladder.rs | 132 --------- .../s0128_longest_consecutive_sequence.rs | 65 ----- .../s0129_sum_root_to_leaf_numbers.rs | 87 ------ src/solution/s0130_surrounded_regions.rs | 221 --------------- src/solution/s0131_palindrome_partitioning.rs | 120 -------- .../s0132_palindrome_partitioning_ii.rs | 91 ------ src/solution/s0134_gas_station.rs | 77 ------ src/solution/s0135_candy.rs | 84 ------ src/solution/s0136_single_number.rs | 48 ---- src/solution/s0137_single_number_ii.rs | 94 ------- src/solution/s0139_word_break.rs | 93 ------- src/solution/s0140_word_break_ii.rs | 72 ----- src/solution/s0143_reorder_list.rs | 50 ---- .../s0144_binary_tree_preorder_traversal.rs | 61 ----- .../s0145_binary_tree_postorder_traversal.rs | 61 ----- src/solution/s0146_lru_cache.rs | 200 -------------- src/solution/s0147_insertion_sort_list.rs | 64 ----- src/solution/s0148_sort_list.rs | 125 --------- src/solution/s0149_max_points_on_a_line.rs | 178 ------------ .../s0150_evaluate_reverse_polish_notation.rs | 91 ------ .../s0151_reverse_words_in_a_string.rs | 93 ------- .../s0152_maximum_product_subarray.rs | 81 ------ ...53_find_minimum_in_rotated_sorted_array.rs | 64 ----- ...find_minimum_in_rotated_sorted_array_ii.rs | 79 ------ src/solution/s0155_min_stack.rs | 116 -------- src/solution/s0162_find_peak_element.rs | 67 ----- src/solution/s0164_maximum_gap.rs | 66 ----- src/solution/s0165_compare_version_numbers.rs | 112 -------- .../s0166_fraction_to_recurring_decimal.rs | 51 ---- .../s0167_two_sum_ii_input_array_is_sorted.rs | 58 ---- .../s0168_excel_sheet_column_title.rs | 78 ------ src/solution/s0169_majority_element.rs | 73 ----- .../s0171_excel_sheet_column_number.rs | 62 ----- .../s0172_factorial_trailing_zeroes.rs | 55 ---- .../s0173_binary_search_tree_iterator.rs | 116 -------- src/solution/s0174_dungeon_game.rs | 115 -------- src/solution/s0179_largest_number.rs | 60 ---- src/solution/s0187_repeated_dna_sequences.rs | 100 ------- ...0188_best_time_to_buy_and_sell_stock_iv.rs | 103 ------- src/solution/s0189_rotate_array.rs | 79 ------ src/solution/s0198_house_robber.rs | 68 ----- .../s0199_binary_tree_right_side_view.rs | 68 ----- 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--git a/README.md b/README.md index 587b18a2..b15ea620 100644 --- a/README.md +++ b/README.md @@ -4,15 +4,9 @@ Run `cargo run {id}` to initialize the template submission file of "question #id Run `cargo test test_{id}` to test the solution for "question #id". -对于大部分难度为 Hard 的问题, 会有中文的思路注释 -Working in progress, to do: -- [ ] auto generation of solution list (when 100 problems solved) +## Ref -## Usage - -* Remove all the solution .rs -* Clean lib.rs file -* Start your leetcode journey in rust by typing `cargo run {question_id}` +[CodeTop](https://codetop.cc/home) diff --git a/src/solution/mod.rs b/src/solution/mod.rs index bc982de9..e69de29b 100644 --- a/src/solution/mod.rs +++ b/src/solution/mod.rs @@ -1,237 +0,0 @@ -mod s0001_two_sum; -mod s0002_add_two_numbers; -mod s0003_longest_substring_without_repeating_characters; -mod s0004_median_of_two_sorted_arrays; -mod s0005_longest_palindromic_substring; -mod s0006_zigzag_conversion; -mod s0007_reverse_integer; -mod s0008_string_to_integer_atoi; -mod s0009_palindrome_number; -mod s0010_regular_expression_matching; -mod s0011_container_with_most_water; -mod s0012_integer_to_roman; -mod s0013_roman_to_integer; -mod s0014_longest_common_prefix; -mod s0015_3sum; -mod s0016_3sum_closest; -mod s0017_letter_combinations_of_a_phone_number; -mod s0018_4sum; -mod s0019_remove_nth_node_from_end_of_list; -mod s0020_valid_parentheses; -mod s0021_merge_two_sorted_lists; -mod s0022_generate_parentheses; -mod s0023_merge_k_sorted_lists; -mod s0024_swap_nodes_in_pairs; -mod s0025_reverse_nodes_in_k_group; -mod s0026_remove_duplicates_from_sorted_array; -mod s0027_remove_element; -mod s0028_implement_strstr; -mod s0029_divide_two_integers; -mod s0030_substring_with_concatenation_of_all_words; -mod s0031_next_permutation; -mod s0032_longest_valid_parentheses; -mod s0033_search_in_rotated_sorted_array; -mod s0034_find_first_and_last_position_of_element_in_sorted_array; -mod s0035_search_insert_position; -mod s0036_valid_sudoku; -mod s0037_sudoku_solver; -mod s0038_count_and_say; -mod s0039_combination_sum; -mod s0040_combination_sum_ii; -mod s0041_first_missing_positive; -mod s0042_trapping_rain_water; -mod s0043_multiply_strings; -mod s0044_wildcard_matching; -mod s0045_jump_game_ii; -mod s0046_permutations; -mod s0047_permutations_ii; -mod s0048_rotate_image; -mod s0049_group_anagrams; -mod s0050_powx_n; -mod s0051_n_queens; -mod s0052_n_queens_ii; -mod s0053_maximum_subarray; -mod s0054_spiral_matrix; -mod s0055_jump_game; -mod s0056_merge_intervals; -mod s0057_insert_interval; -mod s0058_length_of_last_word; -mod s0059_spiral_matrix_ii; -mod s0060_permutation_sequence; -mod s0061_rotate_list; -mod s0062_unique_paths; -mod s0063_unique_paths_ii; -mod s0064_minimum_path_sum; -mod s0065_valid_number; -mod s0066_plus_one; -mod s0067_add_binary; -mod s0068_text_justification; -mod s0069_sqrtx; -mod s0070_climbing_stairs; -mod s0071_simplify_path; -mod s0072_edit_distance; -mod s0073_set_matrix_zeroes; -mod s0074_search_a_2d_matrix; -mod s0075_sort_colors; -mod s0076_minimum_window_substring; -mod s0077_combinations; -mod s0078_subsets; -mod s0079_word_search; -mod s0080_remove_duplicates_from_sorted_array_ii; -mod s0081_search_in_rotated_sorted_array_ii; -mod s0082_remove_duplicates_from_sorted_list_ii; -mod s0083_remove_duplicates_from_sorted_list; -mod s0084_largest_rectangle_in_histogram; -mod s0085_maximal_rectangle; -mod s0086_partition_list; -mod s0087_scramble_string; -mod s0088_merge_sorted_array; -mod s0089_gray_code; -mod s0090_subsets_ii; -mod s0091_decode_ways; -mod s0092_reverse_linked_list_ii; -mod s0093_restore_ip_addresses; -mod s0094_binary_tree_inorder_traversal; -mod s0095_unique_binary_search_trees_ii; -mod s0096_unique_binary_search_trees; -mod s0097_interleaving_string; -mod s0098_validate_binary_search_tree; -mod s0099_recover_binary_search_tree; -mod s0100_same_tree; -mod s0101_symmetric_tree; -mod s0102_binary_tree_level_order_traversal; -mod s0103_binary_tree_zigzag_level_order_traversal; -mod s0104_maximum_depth_of_binary_tree; -mod s0105_construct_binary_tree_from_preorder_and_inorder_traversal; -mod s0106_construct_binary_tree_from_inorder_and_postorder_traversal; -mod s0107_binary_tree_level_order_traversal_ii; -mod s0108_convert_sorted_array_to_binary_search_tree; -mod s0109_convert_sorted_list_to_binary_search_tree; -mod s0110_balanced_binary_tree; -mod s0111_minimum_depth_of_binary_tree; -mod s0112_path_sum; -mod s0113_path_sum_ii; -mod s0114_flatten_binary_tree_to_linked_list; -mod s0115_distinct_subsequences; -mod s0118_pascals_triangle; -mod s0119_pascals_triangle_ii; -mod s0120_triangle; -mod s0121_best_time_to_buy_and_sell_stock; -mod s0122_best_time_to_buy_and_sell_stock_ii; -mod s0123_best_time_to_buy_and_sell_stock_iii; -mod s0124_binary_tree_maximum_path_sum; -mod s0125_valid_palindrome; -mod s0126_word_ladder_ii; -mod s0127_word_ladder; -mod s0128_longest_consecutive_sequence; -mod s0129_sum_root_to_leaf_numbers; -mod s0130_surrounded_regions; -mod s0131_palindrome_partitioning; -mod s0132_palindrome_partitioning_ii; -mod s0134_gas_station; -mod s0135_candy; -mod s0136_single_number; -mod s0137_single_number_ii; -mod s0139_word_break; -mod s0140_word_break_ii; -mod s0143_reorder_list; -mod s0144_binary_tree_preorder_traversal; -mod s0145_binary_tree_postorder_traversal; -mod s0146_lru_cache; -mod s0147_insertion_sort_list; -mod s0148_sort_list; -mod s0149_max_points_on_a_line; -mod s0150_evaluate_reverse_polish_notation; -mod s0151_reverse_words_in_a_string; -mod s0152_maximum_product_subarray; -mod s0153_find_minimum_in_rotated_sorted_array; -mod s0154_find_minimum_in_rotated_sorted_array_ii; -mod s0155_min_stack; -mod s0162_find_peak_element; -mod s0164_maximum_gap; -mod s0165_compare_version_numbers; -mod s0166_fraction_to_recurring_decimal; -mod s0167_two_sum_ii_input_array_is_sorted; -mod s0168_excel_sheet_column_title; -mod s0169_majority_element; -mod s0171_excel_sheet_column_number; -mod s0172_factorial_trailing_zeroes; -mod s0173_binary_search_tree_iterator; -mod s0174_dungeon_game; -mod s0179_largest_number; -mod s0187_repeated_dna_sequences; -mod s0188_best_time_to_buy_and_sell_stock_iv; -mod s0189_rotate_array; -mod s0198_house_robber; -mod s0199_binary_tree_right_side_view; -mod s0200_number_of_islands; -mod s0201_bitwise_and_of_numbers_range; -mod s0202_happy_number; -mod s0203_remove_linked_list_elements; -mod s0204_count_primes; -mod s0205_isomorphic_strings; -mod s0206_reverse_linked_list; -mod s0207_course_schedule; -mod s0208_implement_trie_prefix_tree; -mod s0209_minimum_size_subarray_sum; -mod s0210_course_schedule_ii; -mod s0211_add_and_search_word_data_structure_design; -mod s0212_word_search_ii; -mod s0213_house_robber_ii; -mod s0214_shortest_palindrome; -mod s0215_kth_largest_element_in_an_array; -mod s0216_combination_sum_iii; -mod s0217_contains_duplicate; -mod s0218_the_skyline_problem; -mod s0219_contains_duplicate_ii; -mod s0220_contains_duplicate_iii; -mod s0221_maximal_square; -mod s0222_count_complete_tree_nodes; -mod s0223_rectangle_area; -mod s0224_basic_calculator; -mod s0225_implement_stack_using_queues; -mod s0226_invert_binary_tree; -mod s0227_basic_calculator_ii; -mod s0228_summary_ranges; -mod s0229_majority_element_ii; -mod s0230_kth_smallest_element_in_a_bst; -mod s0231_power_of_two; -mod s0232_implement_queue_using_stacks; -mod s0233_number_of_digit_one; -mod s0238_product_of_array_except_self; -mod s0239_sliding_window_maximum; -mod s0241_different_ways_to_add_parentheses; -mod s0242_valid_anagram; -mod s0257_binary_tree_paths; -mod s0258_add_digits; -mod s0260_single_number_iii; -mod s0263_ugly_number; -mod s0264_ugly_number_ii; -mod s0268_missing_number; -mod s0273_integer_to_english_words; -mod s0274_h_index; -mod s0275_h_index_ii; -mod s0279_perfect_squares; -mod s0282_expression_add_operators; -mod s0283_move_zeroes; -mod s0287_find_the_duplicate_number; -mod s0289_game_of_life; -mod s0290_word_pattern; -mod s0292_nim_game; -mod s0295_find_median_from_data_stream; -mod s0299_bulls_and_cows; -mod s0300_longest_increasing_subsequence; -mod s0301_remove_invalid_parentheses; -mod s0303_range_sum_query_immutable; -mod s0304_range_sum_query_2d_immutable; -mod s0306_additive_number; -mod s0307_range_sum_query_mutable; -mod s0309_best_time_to_buy_and_sell_stock_with_cooldown; -mod s0310_minimum_height_trees; -mod s0312_burst_balloons; -mod s0313_super_ugly_number; -mod s0509_fibonacci_number; -mod s0704_binary_search; -mod s0969_pancake_sorting; -mod s1018_binary_prefix_divisible_by_5; -mod s1046_last_stone_weight; diff --git a/src/solution/s0001_two_sum.rs b/src/solution/s0001_two_sum.rs deleted file mode 100644 index 4a44a732..00000000 --- a/src/solution/s0001_two_sum.rs +++ /dev/null @@ -1,53 +0,0 @@ -/** - * [1] Two Sum - * - * Given an array of integers, return indices of the two numbers such that they - * add up to a specific target. - * - * You may assume that each input would have exactly one solution, and you may - * not use the same element twice. - * - * Example: - * - * - * Given nums = [2, 7, 11, 15], target = 9, - * - * Because nums[0] + nums[1] = 2 + 7 = 9, - * return [0, 1]. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/two-sum/ -// discuss: https://leetcode.com/problems/two-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashMap; -impl Solution { - pub fn two_sum(nums: Vec, target: i32) -> Vec { - let mut map = HashMap::with_capacity(nums.len()); - for (index, num) in nums.iter().enumerate() { - match map.get(&(target - num)) { - None => { - map.insert(num, index); - } - Some(sub_index) => return vec![*sub_index as i32, index as i32], - } - } - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_1() { - assert_eq!(vec![0, 1], Solution::two_sum(vec![2, 7, 11, 15], 9)); - assert_eq!(vec![1, 2], Solution::two_sum(vec![3, 2, 4], 6)); - } -} diff --git a/src/solution/s0002_add_two_numbers.rs b/src/solution/s0002_add_two_numbers.rs deleted file mode 100644 index d8bfc2d8..00000000 --- a/src/solution/s0002_add_two_numbers.rs +++ /dev/null @@ -1,98 +0,0 @@ -/** - * [2] Add Two Numbers - * - * You are given two non-empty linked lists representing two non-negative - * integers. The digits are stored in reverse order and each of their nodes - * contain a single digit. Add the two numbers and return it as a linked list. - * - * You may assume the two numbers do not contain any leading zero, except the - * number 0 itself. - * - * Example: - * - * - * Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) - * Output: 7 -> 0 -> 8 - * Explanation: 342 + 465 = 807. - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/add-two-numbers/ -// discuss: https://leetcode.com/problems/add-two-numbers/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn add_two_numbers( - l1: Option>, - l2: Option>, - ) -> Option> { - let (mut l1, mut l2) = (l1, l2); - let mut dummy_head = Some(Box::new(ListNode::new(0))); - let mut tail = &mut dummy_head; - let (mut l1_end, mut l2_end, mut overflow) = (false, false, false); - loop { - let lhs = match l1 { - Some(node) => { - l1 = node.next; - node.val - } - None => { - l1_end = true; - 0 - } - }; - let rhs = match l2 { - Some(node) => { - l2 = node.next; - node.val - } - None => { - l2_end = true; - 0 - } - }; - // if l1, l2 end and there is not overflow from previous operation, return the result - if l1_end && l2_end && !overflow { - break dummy_head.unwrap().next; - } - let sum = lhs + rhs + if overflow { 1 } else { 0 }; - let sum = if sum >= 10 { - overflow = true; - sum - 10 - } else { - overflow = false; - sum - }; - tail.as_mut().unwrap().next = Some(Box::new(ListNode::new(sum))); - tail = &mut tail.as_mut().unwrap().next - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_2() { - assert_eq!( - Solution::add_two_numbers(to_list(vec![2, 4, 3]), to_list(vec![5, 6, 4])), - to_list(vec![7, 0, 8]) - ); - - assert_eq!( - Solution::add_two_numbers(to_list(vec![9, 9, 9, 9]), to_list(vec![9, 9, 9, 9, 9, 9])), - to_list(vec![8, 9, 9, 9, 0, 0, 1]) - ); - - assert_eq!( - Solution::add_two_numbers(to_list(vec![0]), to_list(vec![0])), - to_list(vec![0]) - ) - } -} diff --git a/src/solution/s0003_longest_substring_without_repeating_characters.rs b/src/solution/s0003_longest_substring_without_repeating_characters.rs deleted file mode 100644 index 0ad6903d..00000000 --- a/src/solution/s0003_longest_substring_without_repeating_characters.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [3] Longest Substring Without Repeating Characters - * - * Given a string, find the length of the longest substring without repeating characters. - * - * Example: - * - * Input: "abcabcbb" - * Output: 3 - * Explanation: The answer is "abc", with the length of 3. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-substring-without-repeating-characters/ -// discuss: https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn length_of_longest_substring(s: String) -> i32 { - let seq: Vec = s.chars().collect(); - let len = seq.len(); - let (mut start, mut end, mut max) = (0, 0, 0); - - while end < len { - for idx in start..end { - if seq[end] == seq[idx] { - start = idx + 1; - break; - } - } - let curr = end - start + 1; - if curr > max { - max = curr - } - end += 1 - } - max as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_3() { - assert_eq!( - Solution::length_of_longest_substring("abcabcbb".to_string()), - 3 - ); - assert_eq!(Solution::length_of_longest_substring("bbbb".to_string()), 1); - assert_eq!( - Solution::length_of_longest_substring("pwwkew".to_string()), - 3 - ); - } -} diff --git a/src/solution/s0004_median_of_two_sorted_arrays.rs b/src/solution/s0004_median_of_two_sorted_arrays.rs deleted file mode 100644 index e3c2d014..00000000 --- a/src/solution/s0004_median_of_two_sorted_arrays.rs +++ /dev/null @@ -1,62 +0,0 @@ -/** - * [4] Median of Two Sorted Arrays - * - * There are two sorted arrays nums1 and nums2 of size m and n respectively. - * - * Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). - * - * You may assume nums1 and nums2 cannot be both empty. - * - * Example 1: - * - * - * nums1 = [1, 3] - * nums2 = [2] - * - * The median is 2.0 - * - * - * Example 2: - * - * - * nums1 = [1, 2] - * nums2 = [3, 4] - * - * The median is (2 + 3)/2 = 2.5 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/median-of-two-sorted-arrays/ -// discuss: https://leetcode.com/problems/median-of-two-sorted-arrays/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: nth slice -impl Solution { - pub fn find_median_sorted_arrays(nums1: Vec, nums2: Vec) -> f64 { - 1.0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - // TODO: implementation - #[test] - #[ignore] - fn test_4() { - assert_eq!( - Solution::find_median_sorted_arrays(vec![1, 3], vec![2]), - 2.0 - ); - assert_eq!( - Solution::find_median_sorted_arrays(vec![1, 2], vec![3, 4]), - 2.5 - ); - } -} diff --git a/src/solution/s0005_longest_palindromic_substring.rs b/src/solution/s0005_longest_palindromic_substring.rs deleted file mode 100644 index 675e1951..00000000 --- a/src/solution/s0005_longest_palindromic_substring.rs +++ /dev/null @@ -1,82 +0,0 @@ -/** - * [5] Longest Palindromic Substring - * - * Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. - * - * Example 1: - * - * - * Input: "babad" - * Output: "bab" - * Note: "aba" is also a valid answer. - * - * - * Example 2: - * - * - * Input: "cbbd" - * Output: "bb" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-palindromic-substring/ -// discuss: https://leetcode.com/problems/longest-palindromic-substring/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn longest_palindrome(s: String) -> String { - let seq: Vec = s.chars().collect(); - let len = seq.len(); - if len < 1 { - return s; - } - let (mut idx, mut curr_len, mut curr_start, mut curr_end) = (0, 0, 0, 0); - while idx < len { - let (mut i, mut j) = (idx, idx); - let ch = seq[idx]; - // handle same char - while i > 0 && seq[i - 1] == ch { - i -= 1 - } - while j < len - 1 && seq[j + 1] == ch { - j += 1 - } - idx = j + 1; - while i > 0 && j < len - 1 && seq[i - 1] == seq[j + 1] { - i -= 1; - j += 1; - } - let max_len = j - i + 1; - if max_len > curr_len { - curr_len = max_len; - curr_start = i; - curr_end = j; - } - if max_len >= len - 1 { - break; - } - } - - s[curr_start..curr_end + 1].to_owned() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_5() { - assert_eq!(Solution::longest_palindrome("aaaaa".to_owned()), "aaaaa"); - assert_eq!(Solution::longest_palindrome("babab".to_owned()), "babab"); - assert_eq!(Solution::longest_palindrome("babcd".to_owned()), "bab"); - assert_eq!(Solution::longest_palindrome("cbbd".to_owned()), "bb"); - assert_eq!(Solution::longest_palindrome("bb".to_owned()), "bb"); - assert_eq!(Solution::longest_palindrome("".to_owned()), ""); - } -} diff --git a/src/solution/s0006_zigzag_conversion.rs b/src/solution/s0006_zigzag_conversion.rs deleted file mode 100644 index 51ed13a7..00000000 --- a/src/solution/s0006_zigzag_conversion.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [6] ZigZag Conversion - * - * The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) - * - * - * P A H N - * A P L S I I G - * Y I R - * - * - * And then read line by line: "PAHNAPLSIIGYIR" - * - * Write the code that will take a string and make this conversion given a number of rows: - * - * - * string convert(string s, int numRows); - * - * Example 1: - * - * - * Input: s = "PAYPALISHIRING", numRows = 3 - * Output: "PAHNAPLSIIGYIR" - * - * - * Example 2: - * - * - * Input: s = "PAYPALISHIRING", numRows = 4 - * Output: "PINALSIGYAHRPI" - * Explanation: - * - * P I N - * A L S I G - * Y A H R - * P I - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/zigzag-conversion/ -// discuss: https://leetcode.com/problems/zigzag-conversion/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn convert(s: String, num_rows: i32) -> String { - if num_rows < 2 { - return s; - } - let seq: Vec = s.chars().collect(); - let len = seq.len(); - let unit: usize = (2 * num_rows - 2) as usize; - let mut sb = String::with_capacity(s.capacity()); - for row in 0..num_rows { - let mut i = row as usize; - let step1 = 2 * (num_rows - 1 - row) as usize; - let step2 = (unit - step1) as usize; - let mut trigger = false; - while i < len { - sb.push(seq[i]); - if step1 == 0 { - i += step2 - } else if step2 == 0 { - i += step1 - } else { - i += if trigger { step2 } else { step1 }; - trigger = !trigger; - } - } - } - sb - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_6() { - assert_eq!( - Solution::convert("PAYPALISHIRING".to_string(), 4), - "PINALSIGYAHRPI" - ); - assert_eq!( - Solution::convert("PAYPALISHIRING".to_string(), 3), - "PAHNAPLSIIGYIR" - ); - assert_eq!(Solution::convert("A".to_string(), 1), "A"); - assert_eq!(Solution::convert("AY".to_string(), 2), "AY"); - } -} diff --git a/src/solution/s0007_reverse_integer.rs b/src/solution/s0007_reverse_integer.rs deleted file mode 100644 index ad89eebd..00000000 --- a/src/solution/s0007_reverse_integer.rs +++ /dev/null @@ -1,73 +0,0 @@ -/** - * [7] Reverse Integer - * - * Given a 32-bit signed integer, reverse digits of an integer. - * - * Example 1: - * - * - * Input: 123 - * Output: 321 - * - * - * Example 2: - * - * - * Input: -123 - * Output: -321 - * - * - * Example 3: - * - * - * Input: 120 - * Output: 21 - * - * - * Note:
- * Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [-2^31, 2^31 - 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/reverse-integer/ -// discuss: https://leetcode.com/problems/reverse-integer/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -impl Solution { - pub fn reverse(x: i32) -> i32 { - let mut input: i64 = x as i64; - let mut result: i64 = 0; - let mut digit: i64 = 0; - let base: i64 = 2; - let upper_bound: i64 = base.pow(31) - 1; - let lower_bound: i64 = -base.pow(31); - while input != 0 { - digit = input % 10; - result = result * 10 + digit; - input = input / 10; - } - if result > upper_bound || result < lower_bound { - return 0; - } - result as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_7() { - assert_eq!(Solution::reverse(123), 321); - assert_eq!(Solution::reverse(-123), -321); - assert_eq!(Solution::reverse(0), 0); - assert_eq!(Solution::reverse(-123000), -321); - let base: i64 = 2; - assert_eq!(Solution::reverse((base.pow(31) - 1) as i32), 0); - assert_eq!(Solution::reverse((-base.pow(31)) as i32), 0); - } -} diff --git a/src/solution/s0008_string_to_integer_atoi.rs b/src/solution/s0008_string_to_integer_atoi.rs deleted file mode 100644 index 9c86e17b..00000000 --- a/src/solution/s0008_string_to_integer_atoi.rs +++ /dev/null @@ -1,131 +0,0 @@ -/** - * [8] String to Integer (atoi) - * - * Implement atoi which converts a string to an integer. - * - * The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value. - * - * The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. - * - * If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed. - * - * If no valid conversion could be performed, a zero value is returned. - * - * Note: - * - *
    - *
  • Only the space character ' ' is considered as whitespace character.
    • - *
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [-2^31, 2^31 - 1]. If the numerical value is out of the range of representable values, INT_MAX (2^31 - 1) or INT_MIN (-2^31) is returned.
    • - *
- * - * Example 1: - * - * - * Input: "42" - * Output: 42 - * - * - * Example 2: - * - * - * Input: " -42" - * Output: -42 - * Explanation: The first non-whitespace character is '-', which is the minus sign. - * Then take as many numerical digits as possible, which gets 42. - * - * - * Example 3: - * - * - * Input: "4193 with words" - * Output: 4193 - * Explanation: Conversion stops at digit '3' as the next character is not a numerical digit. - * - * - * Example 4: - * - * - * Input: "words and 987" - * Output: 0 - * Explanation: The first non-whitespace character is 'w', which is not a numerical - * digit or a +/- sign. Therefore no valid conversion could be performed. - * - * Example 5: - * - * - * Input: "-91283472332" - * Output: -2147483648 - * Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. - * Thefore INT_MIN (-2^31) is returned. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/string-to-integer-atoi/ -// discuss: https://leetcode.com/problems/string-to-integer-atoi/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn my_atoi(input: String) -> i32 { - let (i32_min, i32_max) = (-2_i64.pow(31), 2_i64.pow(31) - 1); - let mut result: i64 = 0; - let mut minus = false; - // simple state machine - let mut num_matched = false; - for ch in input.chars().into_iter() { - if !num_matched { - match ch { - ' ' => {} - '0'..='9' => { - num_matched = true; - result = result * 10 + ch.to_digit(10).unwrap() as i64; - } - '-' => { - num_matched = true; - minus = true; - } - '+' => { - num_matched = true; - } - _ => return 0, - } - } else { - match ch { - '0'..='9' => { - result = result * 10 + ch.to_digit(10).unwrap() as i64; - if result > i32_max { - break; - } - } - _ => break, - } - } - } - result = if minus { -result } else { result }; - if result > i32_max { - return i32_max as i32; - } - if result < i32_min { - return i32_min as i32; - } - return result as i32; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_8() { - assert_eq!(Solution::my_atoi("aa".to_string()), 0); - assert_eq!(Solution::my_atoi("-91283472332".to_string()), -2147483648); - assert_eq!(Solution::my_atoi("words and 987".to_string()), 0); - assert_eq!(Solution::my_atoi("4193 with words".to_string()), 4193); - assert_eq!(Solution::my_atoi("42".to_string()), 42); - assert_eq!(Solution::my_atoi("004193333".to_string()), 4193333); - } -} diff --git a/src/solution/s0009_palindrome_number.rs b/src/solution/s0009_palindrome_number.rs deleted file mode 100644 index 255eeed1..00000000 --- a/src/solution/s0009_palindrome_number.rs +++ /dev/null @@ -1,89 +0,0 @@ -/** - * [9] Palindrome Number - * - * Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. - * - * Example 1: - * - * - * Input: 121 - * Output: true - * - * - * Example 2: - * - * - * Input: -121 - * Output: false - * Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. - * - * - * Example 3: - * - * - * Input: 10 - * Output: false - * Explanation: Reads 01 from right to left. Therefore it is not a palindrome. - * - * - * Follow up: - * - * Coud you solve it without converting the integer to a string? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/palindrome-number/ -// discuss: https://leetcode.com/problems/palindrome-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: not optimal, we only have to revert half of the string -impl Solution { - pub fn is_palindrome(x: i32) -> bool { - if x < 0 { - return false; - } - let mut digits: Vec = Vec::new(); - let mut input = x; - while input != 0 { - digits.push(input % 10); - input = input / 10; - } - let len = digits.len(); - // handle one digit - if len < 2 { - return true; - } - // handle end with 0 - if digits[0] == 0 { - return false; - } - let mut i = 0; - while i < len / 2 { - if digits[i] != digits[len - 1 - i] { - return false; - } - i += 1; - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_9() { - assert_eq!(Solution::is_palindrome(-32), false); - assert_eq!(Solution::is_palindrome(10), false); - assert_eq!(Solution::is_palindrome(0), true); - assert_eq!(Solution::is_palindrome(9), true); - assert_eq!(Solution::is_palindrome(121), true); - assert_eq!(Solution::is_palindrome(2222), true); - assert_eq!(Solution::is_palindrome(11222211), true); - } -} diff --git a/src/solution/s0010_regular_expression_matching.rs b/src/solution/s0010_regular_expression_matching.rs deleted file mode 100644 index 13e15cb7..00000000 --- a/src/solution/s0010_regular_expression_matching.rs +++ /dev/null @@ -1,92 +0,0 @@ -/** - * [10] Regular Expression Matching - * - * Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. - * - * - * '.' Matches any single character. - * '*' Matches zero or more of the preceding element. - * - * - * The matching should cover the entire input string (not partial). - * - * Note: - * - * - * s could be empty and contains only lowercase letters a-z. - * p could be empty and contains only lowercase letters a-z, and characters like . or *. - * - * - * Example 1: - * - * - * Input: - * s = "aa" - * p = "a" - * Output: false - * Explanation: "a" does not match the entire string "aa". - * - * - * Example 2: - * - * - * Input: - * s = "aa" - * p = "a*" - * Output: true - * Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". - * - * - * Example 3: - * - * - * Input: - * s = "ab" - * p = ".*" - * Output: true - * Explanation: ".*" means "zero or more (*) of any character (.)". - * - * - * Example 4: - * - * - * Input: - * s = "aab" - * p = "c*a*b" - * Output: true - * Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab". - * - * - * Example 5: - * - * - * Input: - * s = "mississippi" - * p = "mis*is*p*." - * Output: false - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/regular-expression-matching/ -// discuss: https://leetcode.com/problems/regular-expression-matching/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: NFA -impl Solution { - pub fn is_match(s: String, p: String) -> bool { - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_10() {} -} diff --git a/src/solution/s0011_container_with_most_water.rs b/src/solution/s0011_container_with_most_water.rs deleted file mode 100644 index ae818bac..00000000 --- a/src/solution/s0011_container_with_most_water.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [11] Container With Most Water - * - * Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. - * - * Note: You may not slant the container and n is at least 2. - * - * - * - * - * - * The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49. - * - * - * - * Example: - * - * - * Input: [1,8,6,2,5,4,8,3,7] - * Output: 49 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/container-with-most-water/ -// discuss: https://leetcode.com/problems/container-with-most-water/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Brute force: O(N^2) - -// Two Pointer: a[0] -> <- a[n-1] -impl Solution { - pub fn max_area(height: Vec) -> i32 { - let (mut start, mut end) = (0_usize, (height.len() - 1)); - let mut max: i32 = (end - start) as i32 * Solution::min(height[start], height[end]); - let mut curr_area: i32 = 0; - while end - start > 1 { - // move the lower one - if height[start] < height[end] { - start += 1; - if height[start] < height[start - 1] { - continue; - } - } else { - end -= 1; - if height[end] < height[end + 1] { - continue; - } - } - curr_area = (end - start) as i32 * Solution::min(height[start], height[end]); - if curr_area > max { - max = curr_area - } - } - max - } - - #[inline(always)] - fn min(i: i32, j: i32) -> i32 { - if i > j { - j - } else { - i - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_11() { - assert_eq!(Solution::max_area(vec![1, 8, 6, 2, 5, 4, 8, 3, 7]), 49); - assert_eq!(Solution::max_area(vec![6, 9]), 6); - assert_eq!(Solution::max_area(vec![1, 1, 2, 1, 1, 1]), 5); - } -} diff --git a/src/solution/s0012_integer_to_roman.rs b/src/solution/s0012_integer_to_roman.rs deleted file mode 100644 index b32592c7..00000000 --- a/src/solution/s0012_integer_to_roman.rs +++ /dev/null @@ -1,114 +0,0 @@ -/** - * [12] Integer to Roman - * - * Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. - * - * - * Symbol Value - * I 1 - * V 5 - * X 10 - * L 50 - * C 100 - * D 500 - * M 1000 - * - * For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II. - * - * Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: - * - * - * I can be placed before V (5) and X (10) to make 4 and 9. - * X can be placed before L (50) and C (100) to make 40 and 90. - * C can be placed before D (500) and M (1000) to make 400 and 900. - * - * - * Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999. - * - * Example 1: - * - * - * Input: 3 - * Output: "III" - * - * Example 2: - * - * - * Input: 4 - * Output: "IV" - * - * Example 3: - * - * - * Input: 9 - * Output: "IX" - * - * Example 4: - * - * - * Input: 58 - * Output: "LVIII" - * Explanation: L = 50, V = 5, III = 3. - * - * - * Example 5: - * - * - * Input: 1994 - * Output: "MCMXCIV" - * Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/integer-to-roman/ -// discuss: https://leetcode.com/problems/integer-to-roman/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn int_to_roman(num: i32) -> String { - let table: Vec<(i32, &'static str)> = vec![ - (1000, "M"), - (900, "CM"), - (500, "D"), - (400, "CD"), - (100, "C"), - (90, "XC"), - (50, "L"), - (40, "XL"), - (10, "X"), - (9, "IX"), - (5, "V"), - (4, "IV"), - (1, "I"), - ]; - - let mut num = num; - let mut sb = String::new(); - for p in table.iter() { - if num >= p.0 { - for _ in 0..(num / p.0) { - sb.push_str(p.1); - } - num = num % p.0 - } - } - sb - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_12() { - assert_eq!(Solution::int_to_roman(3), "III"); - assert_eq!(Solution::int_to_roman(4), "IV"); - assert_eq!(Solution::int_to_roman(9), "IX"); - assert_eq!(Solution::int_to_roman(1994), "MCMXCIV"); - } -} diff --git a/src/solution/s0013_roman_to_integer.rs b/src/solution/s0013_roman_to_integer.rs deleted file mode 100644 index 4f3f0ae4..00000000 --- a/src/solution/s0013_roman_to_integer.rs +++ /dev/null @@ -1,115 +0,0 @@ -/** - * [13] Roman to Integer - * - * Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. - * - * - * Symbol Value - * I 1 - * V 5 - * X 10 - * L 50 - * C 100 - * D 500 - * M 1000 - * - * For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II. - * - * Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: - * - * - * I can be placed before V (5) and X (10) to make 4 and 9. - * X can be placed before L (50) and C (100) to make 40 and 90. - * C can be placed before D (500) and M (1000) to make 400 and 900. - * - * - * Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. - * - * Example 1: - * - * - * Input: "III" - * Output: 3 - * - * Example 2: - * - * - * Input: "IV" - * Output: 4 - * - * Example 3: - * - * - * Input: "IX" - * Output: 9 - * - * Example 4: - * - * - * Input: "LVIII" - * Output: 58 - * Explanation: L = 50, V= 5, III = 3. - * - * - * Example 5: - * - * - * Input: "MCMXCIV" - * Output: 1994 - * Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/roman-to-integer/ -// discuss: https://leetcode.com/problems/roman-to-integer/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn roman_to_int(s: String) -> i32 { - let table: Vec<(i32, &'static str)> = vec![ - (1000, "M"), - (900, "CM"), - (500, "D"), - (400, "CD"), - (100, "C"), - (90, "XC"), - (50, "L"), - (40, "XL"), - (10, "X"), - (9, "IX"), - (5, "V"), - (4, "IV"), - (1, "I"), - ]; - let mut sum = 0; - let mut idx = 0; - for p in table.iter() { - while idx + p.1.len() <= s.len() && p.1 == &s[idx..idx + p.1.len()] { - idx += p.1.len(); - sum += p.0; - if idx >= s.len() { - return sum; - } - } - } - sum - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_13() { - assert_eq!(Solution::roman_to_int("III".to_string()), 3); - assert_eq!(Solution::roman_to_int("IV".to_string()), 4); - assert_eq!(Solution::roman_to_int("IX".to_string()), 9); - assert_eq!(Solution::roman_to_int("MCMXCIV".to_string()), 1994); - assert_eq!(Solution::roman_to_int("DCXXI".to_string()), 621); - } -} diff --git a/src/solution/s0014_longest_common_prefix.rs b/src/solution/s0014_longest_common_prefix.rs deleted file mode 100644 index 1c5fd985..00000000 --- a/src/solution/s0014_longest_common_prefix.rs +++ /dev/null @@ -1,90 +0,0 @@ -/** - * [14] Longest Common Prefix - * - * Write a function to find the longest common prefix string amongst an array of strings. - * - * If there is no common prefix, return an empty string "". - * - * Example 1: - * - * - * Input: ["flower","flow","flight"] - * Output: "fl" - * - * - * Example 2: - * - * - * Input: ["dog","racecar","car"] - * Output: "" - * Explanation: There is no common prefix among the input strings. - * - * - * Note: - * - * All given inputs are in lowercase letters a-z. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-common-prefix/ -// discuss: https://leetcode.com/problems/longest-common-prefix/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::str::Chars; -impl Solution { - pub fn longest_common_prefix(strs: Vec) -> String { - let mut prefix = String::new(); - let mut iters: Vec = strs.iter().map(|s| s.chars()).collect(); - let mut curr_char: Option = None; - if strs.len() < 1 { - return prefix; - } - loop { - curr_char.take().map(|ch| prefix.push(ch)); - for iter in iters.iter_mut() { - let mut ch = iter.next(); - if ch.is_none() { - return prefix; - } - match curr_char { - None => curr_char = ch.take(), - Some(curr) => { - if curr != ch.unwrap() { - return prefix; - } - } - } - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_14() { - assert_eq!( - Solution::longest_common_prefix(vec![ - "".to_string(), - "racecar".to_string(), - "car".to_string() - ]), - "" - ); - assert_eq!( - Solution::longest_common_prefix(vec![ - "flower".to_string(), - "flow".to_string(), - "flight".to_string() - ]), - "fl" - ); - assert_eq!(Solution::longest_common_prefix(vec![]), ""); - } -} diff --git a/src/solution/s0015_3sum.rs b/src/solution/s0015_3sum.rs deleted file mode 100644 index 9feb3cd0..00000000 --- a/src/solution/s0015_3sum.rs +++ /dev/null @@ -1,143 +0,0 @@ -/** - * [15] 3Sum - * - * Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. - * - * Note: - * - * The solution set must not contain duplicate triplets. - * - * Example: - * - * - * Given array nums = [-1, 0, 1, 2, -1, -4], - * - * A solution set is: - * [ - * [-1, 0, 1], - * [-1, -1, 2] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/3sum/ -// discuss: https://leetcode.com/problems/3sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn three_sum(nums: Vec) -> Vec> { - let len = nums.len(); - if len < 3 { - return vec![]; - } - let mut nums = nums; - nums.sort(); - let mut i = 0; - let mut result: Vec> = Vec::new(); - let mut previous = nums[0] - 1; - while i < len - 2 { - // skip same number - if nums[i] == previous { - i += 1; - continue; - } - previous = nums[i]; - let mut vec = Solution::two_sum(&nums[(i + 1)..len], 0 - nums[i]); - for t in vec.iter() { - result.push(vec![nums[i], t.0, t.1]); - } - i += 1; - } - result - } - - // 2 sum using 2 pointers: nums[0] -> <- nums[len-1] - #[inline(always)] - fn two_sum(nums: &[i32], sum: i32) -> Vec<(i32, i32)> { - let (mut i, mut j) = (0_usize, nums.len() - 1); - let mut result = Vec::new(); - while i < j { - if nums[i] + nums[j] < sum { - i += 1 - } else if nums[i] + nums[j] > sum { - j -= 1 - } else { - result.push((nums[i], nums[j])); - i = Solution::next_unique(nums, i, true); - j = Solution::next_unique(nums, j, false); - } - } - result - } - - // seek next un-repeat number - #[inline(always)] - fn next_unique(nums: &[i32], idx: usize, forward: bool) -> usize { - let curr = nums[idx]; - let mut i = idx; - while i > 0 && i < nums.len() && nums[i] == curr { - i = if forward { i + 1 } else { i - 1 } - } - i - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_15() { - assert_eq!( - Solution::three_sum(vec![-1, 0, 1, 2, -1, -4]), - vec![vec![-1, -1, 2], vec![-1, 0, 1]] - ); - assert_eq!( - Solution::three_sum(vec![ - -7, -4, -6, 6, 4, -6, -9, -10, -7, 5, 3, -1, -5, 8, -1, -2, -8, -1, 5, -3, -5, 4, - 2, -5, -4, 4, 7 - ]), - vec![ - vec![-10, 2, 8], - vec![-10, 3, 7], - vec![-10, 4, 6], - vec![-10, 5, 5], - vec![-9, 2, 7], - vec![-9, 3, 6], - vec![-9, 4, 5], - vec![-8, 2, 6], - vec![-8, 3, 5], - vec![-8, 4, 4], - vec![-7, -1, 8], - vec![-7, 2, 5], - vec![-7, 3, 4], - vec![-6, -2, 8], - vec![-6, -1, 7], - vec![-6, 2, 4], - vec![-5, -3, 8], - vec![-5, -2, 7], - vec![-5, -1, 6], - vec![-5, 2, 3], - vec![-4, -4, 8], - vec![-4, -3, 7], - vec![-4, -2, 6], - vec![-4, -1, 5], - vec![-3, -2, 5], - vec![-3, -1, 4], - vec![-2, -1, 3], - vec![-1, -1, 2] - ] - ); - assert_eq!( - Solution::three_sum(vec![2, 0, -2, -5, -5, -3, 2, -4]), - vec![vec![-4, 2, 2], vec![-2, 0, 2]] - ); - let empty_vec: Vec> = vec![]; - assert_eq!(Solution::three_sum(vec![]), empty_vec); - } -} diff --git a/src/solution/s0016_3sum_closest.rs b/src/solution/s0016_3sum_closest.rs deleted file mode 100644 index 01cf442d..00000000 --- a/src/solution/s0016_3sum_closest.rs +++ /dev/null @@ -1,76 +0,0 @@ -/** - * [16] 3Sum Closest - * - * Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution. - * - * Example: - * - * - * Given array nums = [-1, 2, 1, -4], and target = 1. - * - * The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/3sum-closest/ -// discuss: https://leetcode.com/problems/3sum-closest/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn three_sum_closest(nums: Vec, target: i32) -> i32 { - let mut nums = nums; - let mut min_distance: i32 = i32::max_value(); - nums.sort(); - let mut i = 0; - while i < nums.len() - 2 { - let sub_min = Solution::two_sum_closest(&nums[(i + 1)..nums.len()], target - nums[i]); - if sub_min.abs() < min_distance.abs() { - min_distance = sub_min; - if min_distance == 0 { - break; - } - } - i += 1; - } - target + min_distance - } - - pub fn two_sum_closest(nums: &[i32], target: i32) -> i32 { - let (mut i, mut j) = (0_usize, nums.len() - 1); - let mut local_min = i32::max_value(); - while i < j { - let sum = nums[i] + nums[j]; - if sum > target { - j -= 1; - } else if sum < target { - i += 1; - } else { - return 0; - } - if (sum - target).abs() < local_min.abs() { - local_min = sum - target - } - } - local_min - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_16() { - assert_eq!(Solution::three_sum_closest(vec![-1, 2, 1, -4], 1), 2); - assert_eq!(Solution::three_sum_closest(vec![1, 2, 3], 1), 6); - assert_eq!( - Solution::three_sum_closest(vec![1, 2, 4, 8, 16, 32, 64, 128], 82), - 82 - ); - } -} diff --git a/src/solution/s0017_letter_combinations_of_a_phone_number.rs b/src/solution/s0017_letter_combinations_of_a_phone_number.rs deleted file mode 100644 index 7d3245f4..00000000 --- a/src/solution/s0017_letter_combinations_of_a_phone_number.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [17] Letter Combinations of a Phone Number - * - * Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. - * - * A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters. - * - * - * - * Example: - * - * - * Input: "23" - * Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. - * - * - * Note: - * - * Although the above answer is in lexicographical order, your answer could be in any order you want. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/letter-combinations-of-a-phone-number/ -// discuss: https://leetcode.com/problems/letter-combinations-of-a-phone-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn letter_combinations(digits: String) -> Vec { - // '0' and '1' as placeholder to avoid index shifting - let table: Vec<(char, Vec)> = vec![ - ('0', vec![]), - ('1', vec![]), - ('2', vec!['a', 'b', 'c']), - ('3', vec!['d', 'e', 'f']), - ('4', vec!['g', 'h', 'i']), - ('5', vec!['j', 'k', 'l']), - ('6', vec!['m', 'n', 'o']), - ('7', vec!['p', 'q', 'r', 's']), - ('8', vec!['t', 'u', 'v']), - ('9', vec!['w', 'x', 'y', 'z']), - ]; - if digits.len() < 1 { - return vec![]; - } - let mut combs: Vec = vec![String::with_capacity(digits.len())]; - for ch in digits.chars().into_iter() { - let chs = &table[ch.to_digit(10).unwrap() as usize].1; - let mut added: Vec = Vec::with_capacity((chs.len() - 1) * combs.len()); - for comb in combs.iter_mut() { - for (i, &alphabetic) in chs.iter().enumerate() { - if i == chs.len() - 1 { - comb.push(alphabetic); - } else { - let mut new_comb = (*comb).clone(); - new_comb.push(alphabetic); - added.push(new_comb); - } - } - } - combs.append(&mut added); - } - combs - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_17() { - assert_eq!( - Solution::letter_combinations("23".to_string()), - ["cf", "af", "bf", "cd", "ce", "ad", "ae", "bd", "be"] - ); - } -} diff --git a/src/solution/s0018_4sum.rs b/src/solution/s0018_4sum.rs deleted file mode 100644 index d832aef1..00000000 --- a/src/solution/s0018_4sum.rs +++ /dev/null @@ -1,99 +0,0 @@ -/** - * [18] 4Sum - * - * Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. - * - * Note: - * - * The solution set must not contain duplicate quadruplets. - * - * Example: - * - * - * Given array nums = [1, 0, -1, 0, -2, 2], and target = 0. - * - * A solution set is: - * [ - * [-1, 0, 0, 1], - * [-2, -1, 1, 2], - * [-2, 0, 0, 2] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/4sum/ -// discuss: https://leetcode.com/problems/4sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: change to faster N^3 solution... maybe -// this is a N^2 * logN solution, but slower than N^3 solution -// iterate all combinations and the sum of 2 elements, then use one-round hash -use std::collections::BTreeMap; -use std::collections::HashSet; -impl Solution { - pub fn four_sum(nums: Vec, target: i32) -> Vec> { - if nums.len() < 4 { - return vec![]; - } - let mut set: HashSet> = HashSet::new(); - let mut map: BTreeMap> = BTreeMap::new(); - // collect two-sums in asc order, store the index to avoid single number reusing - for i in 0..(nums.len() - 1) { - for j in (i + 1)..nums.len() { - map.entry(nums[i] + nums[j]) - .or_insert(Vec::new()) - .push((i, j)); - } - } - // find results - for (&sum, pairs) in map.iter() { - // avoid duplicates - if sum > target / 2 { - break; - } - match map.get(&(target - sum)) { - None => continue, - // 2-sum + 2-sum == target, then all the possible combination - // (without index conflicts) is our answer - Some(subs) => { - for pair in pairs.iter() { - for sub in subs.iter() { - if sub.0 == pair.0 - || sub.0 == pair.1 - || sub.1 == pair.0 - || sub.1 == pair.1 - { - continue; - } - let mut vec = - vec![nums[pair.0], nums[pair.1], nums[sub.0], nums[sub.1]]; - vec.sort(); - set.insert(vec); - } - } - } - } - } - set.into_iter().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - // TODO: build a macro for arbitrary match - #[test] - #[ignore] - fn test_18() { - assert_eq!( - Solution::four_sum(vec![1, 0, -1, 0, -2, 2], 0), - vec![vec![-1, 0, 0, 1], vec![-2, 0, 0, 2], vec![-2, -1, 1, 2]] - ); - } -} diff --git a/src/solution/s0019_remove_nth_node_from_end_of_list.rs b/src/solution/s0019_remove_nth_node_from_end_of_list.rs deleted file mode 100644 index c3f53a32..00000000 --- a/src/solution/s0019_remove_nth_node_from_end_of_list.rs +++ /dev/null @@ -1,71 +0,0 @@ -/** - * [19] Remove Nth Node From End of List - * - * Given a linked list, remove the n-th node from the end of list and return its head. - * - * Example: - * - * - * Given linked list: 1->2->3->4->5, and n = 2. - * - * After removing the second node from the end, the linked list becomes 1->2->3->5. - * - * - * Note: - * - * Given n will always be valid. - * - * Follow up: - * - * Could you do this in one pass? - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/remove-nth-node-from-end-of-list/ -// discuss: https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// one pass (two pointer runner pattern) cannot make borrow checker happy -// but two pass don't takes longer time -impl Solution { - pub fn remove_nth_from_end(head: Option>, n: i32) -> Option> { - let mut dummy_head = Some(Box::new(ListNode { val: 0, next: head })); - let mut len = 0; - { - let mut p = dummy_head.as_ref(); - while p.unwrap().next.is_some() { - len += 1; - p = p.unwrap().next.as_ref(); - } - } - let idx = len - n; - { - let mut p = dummy_head.as_mut(); - for _ in 0..(idx) { - p = p.unwrap().next.as_mut(); - } - let next = p.as_mut().unwrap().next.as_mut().unwrap().next.take(); - p.as_mut().unwrap().next = next; - } - dummy_head.unwrap().next - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_19() { - assert_eq!( - Solution::remove_nth_from_end(to_list(vec![1, 2, 3, 4, 5]), 2), - to_list(vec![1, 2, 3, 5]) - ); - assert_eq!(Solution::remove_nth_from_end(to_list(vec![1]), 1), None); - } -} diff --git a/src/solution/s0020_valid_parentheses.rs b/src/solution/s0020_valid_parentheses.rs deleted file mode 100644 index 3c0940b9..00000000 --- a/src/solution/s0020_valid_parentheses.rs +++ /dev/null @@ -1,94 +0,0 @@ -/** - * [20] Valid Parentheses - * - * Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. - * - * An input string is valid if: - * - *
    - * Open brackets must be closed by the same type of brackets. - * Open brackets must be closed in the correct order. - *
- * - * Note that an empty string is also considered valid. - * - * Example 1: - * - * - * Input: "()" - * Output: true - * - * - * Example 2: - * - * - * Input: "()[]{}" - * Output: true - * - * - * Example 3: - * - * - * Input: "(]" - * Output: false - * - * - * Example 4: - * - * - * Input: "([)]" - * Output: false - * - * - * Example 5: - * - * - * Input: "{[]}" - * Output: true - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/valid-parentheses/ -// discuss: https://leetcode.com/problems/valid-parentheses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_valid(s: String) -> bool { - let mut stack: Vec = Vec::new(); - for ch in s.chars().into_iter() { - match stack.last() { - None => {} - Some(&last) => { - if Solution::pair(last, ch) { - stack.pop(); - continue; - } - } - } - stack.push(ch); - } - stack.is_empty() - } - - #[inline(always)] - fn pair(open: char, close: char) -> bool { - (open == '{' && close == '}') - || (open == '(' && close == ')') - || (open == '[' && close == ']') - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_20() { - assert_eq!(Solution::is_valid("()[]{}".to_string()), true); - } -} diff --git a/src/solution/s0021_merge_two_sorted_lists.rs b/src/solution/s0021_merge_two_sorted_lists.rs deleted file mode 100644 index 54d7f256..00000000 --- a/src/solution/s0021_merge_two_sorted_lists.rs +++ /dev/null @@ -1,75 +0,0 @@ -/** - * [21] Merge Two Sorted Lists - * - * Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. - * - * Example: - * - * Input: 1->2->4, 1->3->4 - * Output: 1->1->2->3->4->4 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/merge-two-sorted-lists/ -// discuss: https://leetcode.com/problems/merge-two-sorted-lists/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// recursive will be much easier to understand -impl Solution { - pub fn merge_two_lists( - l1: Option>, - l2: Option>, - ) -> Option> { - let mut dummy_head = Some(Box::new(ListNode { val: 0, next: None })); - let mut head = &mut dummy_head; - let (mut l1, mut l2) = (l1, l2); - while l1.is_some() || l2.is_some() { - if l1.is_none() { - head.as_mut().unwrap().next = l2; - break; - } else if l2.is_none() { - head.as_mut().unwrap().next = l1; - break; - } - let next = if l1.as_ref().unwrap().val < l2.as_ref().unwrap().val { - let (origin, next) = Solution::take_head(l1); - l1 = origin; - next - } else { - let (origin, next) = Solution::take_head(l2); - l2 = origin; - next - }; - head.as_mut().unwrap().next = next; - head = &mut head.as_mut().unwrap().next; - } - dummy_head.unwrap().next - } - - #[inline(always)] - fn take_head(mut l: Option>) -> (Option>, Option>) { - let l_next = l.as_mut().unwrap().next.take(); - let next = l.take(); - l = l_next; - (l, next) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_21() { - assert_eq!( - Solution::merge_two_lists(to_list(vec![1, 2, 4]), to_list(vec![1, 3, 4])), - to_list(vec![1, 1, 2, 3, 4, 4]) - ); - } -} diff --git a/src/solution/s0022_generate_parentheses.rs b/src/solution/s0022_generate_parentheses.rs deleted file mode 100644 index 9e0e33df..00000000 --- a/src/solution/s0022_generate_parentheses.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [22] Generate Parentheses - * - * - * Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. - * - * - * - * For example, given n = 3, a solution set is: - * - * - * [ - * "((()))", - * "(()())", - * "(())()", - * "()(())", - * "()()()" - * ] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/generate-parentheses/ -// discuss: https://leetcode.com/problems/generate-parentheses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// DFS -impl Solution { - pub fn generate_parenthesis(n: i32) -> Vec { - if n < 1 { - return vec![]; - } - let mut result = Vec::new(); - Solution::dfs(n, 0, 0, &mut result, String::new()); - result - } - - fn dfs(n: i32, left: i32, right: i32, result: &mut Vec, mut path: String) { - if left == n && right == n { - result.push(path); - return; - } - if left < n { - let mut new_path = path.clone(); - new_path.push('('); - Solution::dfs(n, left + 1, right, result, new_path); - } - if right < left { - // reuse path to avoid clone overhead - path.push(')'); - Solution::dfs(n, left, right + 1, result, path); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_22() { - assert_eq!(Solution::generate_parenthesis(1), vec!["()"]); - assert_eq!(Solution::generate_parenthesis(2), vec!["(())", "()()"]); - assert_eq!( - Solution::generate_parenthesis(3), - vec!["((()))", "(()())", "(())()", "()(())", "()()()"] - ); - } -} diff --git a/src/solution/s0023_merge_k_sorted_lists.rs b/src/solution/s0023_merge_k_sorted_lists.rs deleted file mode 100644 index eeaf3695..00000000 --- a/src/solution/s0023_merge_k_sorted_lists.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [23] Merge k Sorted Lists - * - * Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. - * - * Example: - * - * - * Input: - * [ - * 1->4->5, - * 1->3->4, - * 2->6 - * ] - * Output: 1->1->2->3->4->4->5->6 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/merge-k-sorted-lists/ -// discuss: https://leetcode.com/problems/merge-k-sorted-lists/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -use std::cmp::Ordering; -use std::collections::BinaryHeap; - -// head value and the index -struct Node(i32, usize); - -// sort in reverse order of head value -impl Ord for Node { - fn cmp(&self, other: &Self) -> Ordering { - self.0.cmp(&other.0).reverse() - } -} -impl PartialEq for Node { - fn eq(&self, other: &Self) -> bool { - self.0.eq(&other.0) - } -} -impl Eq for Node {} -impl PartialOrd for Node { - fn partial_cmp(&self, other: &Self) -> Option { - Some(self.0.cmp(&other.0).reverse()) - } -} - -impl Solution { - pub fn merge_k_lists(lists: Vec>>) -> Option> { - let mut heap: BinaryHeap = BinaryHeap::new(); - for (idx, node) in lists.iter().enumerate() { - node.as_ref() - .and_then(|n| Some(heap.push(Node(n.val, idx)))); - } - Solution::next(lists, &mut heap) - } - - fn next( - mut lists: Vec>>, - heap: &mut BinaryHeap, - ) -> Option> { - heap.pop().map(|node| { - let next = lists[node.1].take().unwrap().next; - next.as_ref() - .and_then(|n| Some(heap.push(Node(n.val, node.1)))); - lists[node.1] = next; - Box::new(ListNode { - val: node.0, - next: Solution::next(lists, heap), - }) - }) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_23() { - assert_eq!( - Solution::merge_k_lists(vec![ - to_list(vec![1, 4, 5]), - to_list(vec![1, 3, 4]), - to_list(vec![2, 6]), - ]), - to_list(vec![1, 1, 2, 3, 4, 4, 5, 6]) - ); - assert_eq!(Solution::merge_k_lists(vec![]), None); - } -} diff --git a/src/solution/s0024_swap_nodes_in_pairs.rs b/src/solution/s0024_swap_nodes_in_pairs.rs deleted file mode 100644 index 4a6deb3e..00000000 --- a/src/solution/s0024_swap_nodes_in_pairs.rs +++ /dev/null @@ -1,73 +0,0 @@ -/** - * [24] Swap Nodes in Pairs - * - * Given a linked list, swap every two adjacent nodes and return its head. - * - * Example: - * - * - * Given 1->2->3->4, you should return the list as 2->1->4->3. - * - * Note: - * - * - * Your algorithm should use only constant extra space. - * You may not modify the values in the list's nodes, only nodes itself may be changed. - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/swap-nodes-in-pairs/ -// discuss: https://leetcode.com/problems/swap-nodes-in-pairs/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn swap_pairs(head: Option>) -> Option> { - let mut dummy_head = Some(Box::new(ListNode { val: 0, next: head })); - let mut head = dummy_head.as_mut(); - loop { - let mut left = head.as_mut().unwrap().next.take(); - if left.is_none() { - break; - } - let mut right = left.as_mut().unwrap().next.take(); - // handle the un-paired one, e.g. [1, 2, 3] -> [2, 1, 3], 3 is un-paired - if right.is_none() { - head.as_mut().unwrap().next = left; - break; - } - let mut next = right.as_mut().unwrap().next.take(); - // BEFORE: head -> left -> right -> next - // AFTER: head -> right -> left -> next - left.as_mut().unwrap().next = next; - right.as_mut().unwrap().next = left; - head.as_mut().unwrap().next = right; - head = head.unwrap().next.as_mut().unwrap().next.as_mut(); - } - dummy_head.unwrap().next - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_24() { - assert_eq!( - Solution::swap_pairs(to_list(vec![1, 2, 3, 4])), - to_list(vec![2, 1, 4, 3]) - ); - assert_eq!(Solution::swap_pairs(to_list(vec![])), to_list(vec![])); - assert_eq!( - Solution::swap_pairs(to_list(vec![1, 2, 3])), - to_list(vec![2, 1, 3]) - ); - assert_eq!(Solution::swap_pairs(to_list(vec![1])), to_list(vec![1])); - } -} diff --git a/src/solution/s0025_reverse_nodes_in_k_group.rs b/src/solution/s0025_reverse_nodes_in_k_group.rs deleted file mode 100644 index 5d39cd1e..00000000 --- a/src/solution/s0025_reverse_nodes_in_k_group.rs +++ /dev/null @@ -1,106 +0,0 @@ -/** - * [25] Reverse Nodes in k-Group - * - * Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. - * - * k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. - * - * - * - * - * Example: - * - * Given this linked list: 1->2->3->4->5 - * - * For k = 2, you should return: 2->1->4->3->5 - * - * For k = 3, you should return: 3->2->1->4->5 - * - * Note: - * - * - * Only constant extra memory is allowed. - * You may not alter the values in the list's nodes, only nodes itself may be changed. - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/reverse-nodes-in-k-group/ -// discuss: https://leetcode.com/problems/reverse-nodes-in-k-group/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn reverse_k_group(head: Option>, k: i32) -> Option> { - let mut dummy_head = Some(Box::new(ListNode { val: 0, next: head })); - let mut head = dummy_head.as_mut(); - 'outer: loop { - let mut start = head.as_mut().unwrap().next.take(); - if start.is_none() { - break 'outer; - } - let mut end = start.as_mut(); - for _ in 0..(k - 1) { - end = end.unwrap().next.as_mut(); - if end.is_none() { - head.as_mut().unwrap().next = start; - break 'outer; - } - } - let mut tail = end.as_mut().unwrap().next.take(); - // BEFORE: head -> start -> 123456... -> end -> tail - // AFTER: head -> end -> ...654321 -> start -> tail - let end = Solution::reverse(start, tail); - head.as_mut().unwrap().next = end; - for _ in 0..k { - head = head.unwrap().next.as_mut() - } - } - dummy_head.unwrap().next - } - - #[inline(always)] - fn reverse( - mut head: Option>, - tail: Option>, - ) -> Option> { - let mut prev = tail; - let mut current = head; - while let Some(mut current_node_inner) = current { - let mut next = current_node_inner.next.take(); - current_node_inner.next = prev.take(); - prev = Some(current_node_inner); - current = next; - } - prev - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_25() { - assert_eq!( - Solution::reverse_k_group(to_list(vec![1, 2, 3, 4, 5]), 2), - to_list(vec![2, 1, 4, 3, 5]) - ); - assert_eq!( - Solution::reverse_k_group(to_list(vec![1, 2, 3, 4, 5]), 3), - to_list(vec![3, 2, 1, 4, 5]) - ); - assert_eq!( - Solution::reverse_k_group(to_list(vec![1, 2, 3, 4, 5]), 5), - to_list(vec![5, 4, 3, 2, 1]) - ); - assert_eq!( - Solution::reverse_k_group(to_list(vec![1]), 1), - to_list(vec![1]) - ); - } -} diff --git a/src/solution/s0026_remove_duplicates_from_sorted_array.rs b/src/solution/s0026_remove_duplicates_from_sorted_array.rs deleted file mode 100644 index 286b7c54..00000000 --- a/src/solution/s0026_remove_duplicates_from_sorted_array.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [26] Remove Duplicates from Sorted Array - * - * Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. - * - * Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. - * - * Example 1: - * - * - * Given nums = [1,1,2], - * - * Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. - * - * It doesn't matter what you leave beyond the returned length. - * - * Example 2: - * - * - * Given nums = [0,0,1,1,1,2,2,3,3,4], - * - * Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. - * - * It doesn't matter what values are set beyond the returned length. - * - * - * Clarification: - * - * Confused why the returned value is an integer but your answer is an array? - * - * Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. - * - * Internally you can think of this: - * - * - * // nums is passed in by reference. (i.e., without making a copy) - * int len = removeDuplicates(nums); - * - * // any modification to nums in your function would be known by the caller. - * // using the length returned by your function, it prints the first len elements. - * for (int i = 0; i < len; i++) { - * print(nums[i]); - * } - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/remove-duplicates-from-sorted-array/ -// discuss: https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn remove_duplicates(nums: &mut Vec) -> i32 { - let len = nums.len(); - if len <= 1 { - return len as i32; - } - let mut slow = 0usize; - for fast in 1..len { - if nums[slow] != nums[fast] { - slow += 1; - nums[slow] = nums[fast]; - } - } - nums.truncate(slow + 1); - (slow + 1) as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_26() { - assert_eq!(Solution::remove_duplicates(&mut vec![]), 0); - let mut vec1 = vec![1, 1, 1, 1, 3]; - assert_eq!(Solution::remove_duplicates(&mut vec1), 2); - assert_eq!(vec1, vec![1, 3]); - let mut vec2 = vec![1, 1, 2]; - assert_eq!(Solution::remove_duplicates(&mut vec2), 2); - assert_eq!(vec2, vec![1, 2]); - } -} diff --git a/src/solution/s0027_remove_element.rs b/src/solution/s0027_remove_element.rs deleted file mode 100644 index 7144d3e9..00000000 --- a/src/solution/s0027_remove_element.rs +++ /dev/null @@ -1,97 +0,0 @@ -/** - * [27] Remove Element - * - * Given an array nums and a value val, remove all instances of that value in-place and return the new length. - * - * Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. - * - * The order of elements can be changed. It doesn't matter what you leave beyond the new length. - * - * Example 1: - * - * - * Given nums = [3,2,2,3], val = 3, - * - * Your function should return length = 2, with the first two elements of nums being 2. - * - * It doesn't matter what you leave beyond the returned length. - * - * - * Example 2: - * - * - * Given nums = [0,1,2,2,3,0,4,2], val = 2, - * - * Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4. - * - * Note that the order of those five elements can be arbitrary. - * - * It doesn't matter what values are set beyond the returned length. - * - * Clarification: - * - * Confused why the returned value is an integer but your answer is an array? - * - * Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. - * - * Internally you can think of this: - * - * - * // nums is passed in by reference. (i.e., without making a copy) - * int len = removeElement(nums, val); - * - * // any modification to nums in your function would be known by the caller. - * // using the length returned by your function, it prints the first len elements. - * for (int i = 0; i < len; i++) { - * print(nums[i]); - * } - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/remove-element/ -// discuss: https://leetcode.com/problems/remove-element/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn remove_element(nums: &mut Vec, val: i32) -> i32 { - if nums.len() < 1 { - return 0; - } - let (mut start, mut end) = (0_usize, nums.len() - 1); - while start < end { - if nums[start] == val { - nums[start] = nums[end - 1]; - end -= 1; - } else { - start += 1; - } - } - end as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_27() { - let mut vec1 = vec![0, 1, 2, 2, 3, 0, 4, 2]; - assert_eq!(Solution::remove_element(&mut vec1, 2), 5); - assert_eq!(vec1[0..5], [0, 1, 4, 0, 3]); - assert_eq!(Solution::remove_element(&mut vec![], 2), 0); - assert_eq!( - Solution::remove_element(&mut vec![1, 2, 2, 2, 2, 2, 2], 2), - 1 - ); - assert_eq!( - Solution::remove_element(&mut vec![2, 2, 2, 2, 2, 2, 2], 2), - 0 - ); - assert_eq!(Solution::remove_element(&mut vec![1], 1), 0); - } -} diff --git a/src/solution/s0028_implement_strstr.rs b/src/solution/s0028_implement_strstr.rs deleted file mode 100644 index e0d944ee..00000000 --- a/src/solution/s0028_implement_strstr.rs +++ /dev/null @@ -1,55 +0,0 @@ -/** - * [28] Implement strStr() - * - * Implement strStr(). - * - * Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. - * - * Example 1: - * - * - * Input: haystack = "hello", needle = "ll" - * Output: 2 - * - * - * Example 2: - * - * - * Input: haystack = "aaaaa", needle = "bba" - * Output: -1 - * - * - * Clarification: - * - * What should we return when needle is an empty string? This is a great question to ask during an interview. - * - * For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf(). - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/implement-strstr/ -// discuss: https://leetcode.com/problems/implement-strstr/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn str_str(haystack: String, needle: String) -> i32 { - if needle.is_empty() { - return 0; - } - haystack.find(&needle).map_or(-1_i32, |v| v as i32) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_28() { - assert_eq!(Solution::str_str("hello".to_string(), "ll".to_string()), 2) - } -} diff --git a/src/solution/s0029_divide_two_integers.rs b/src/solution/s0029_divide_two_integers.rs deleted file mode 100644 index e8f13435..00000000 --- a/src/solution/s0029_divide_two_integers.rs +++ /dev/null @@ -1,52 +0,0 @@ -/** - * [29] Divide Two Integers - * - * Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. - * - * Return the quotient after dividing dividend by divisor. - * - * The integer division should truncate toward zero. - * - * Example 1: - * - * - * Input: dividend = 10, divisor = 3 - * Output: 3 - * - * Example 2: - * - * - * Input: dividend = 7, divisor = -3 - * Output: -2 - * - * Note: - * - * - * Both dividend and divisor will be 32-bit signed integers. - * The divisor will never be 0. - * Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [-2^31, 2^31 - 1]. For the purpose of this problem, assume that your function returns 2^31 - 1 when the division result overflows. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/divide-two-integers/ -// discuss: https://leetcode.com/problems/divide-two-integers/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn divide(dividend: i32, divisor: i32) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_29() {} -} diff --git a/src/solution/s0030_substring_with_concatenation_of_all_words.rs b/src/solution/s0030_substring_with_concatenation_of_all_words.rs deleted file mode 100644 index 002ae9a3..00000000 --- a/src/solution/s0030_substring_with_concatenation_of_all_words.rs +++ /dev/null @@ -1,175 +0,0 @@ -/** - * [30] Substring with Concatenation of All Words - * - * You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters. - * - * Example 1: - * - * - * Input: - * s = "barfoothefoobarman", - * words = ["foo","bar"] - * Output: [0,9] - * Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively. - * The output order does not matter, returning [9,0] is fine too. - * - * - * Example 2: - * - * - * Input: - * s = "wordgoodgoodgoodbestword", - * words = ["word","good","best","word"] - * Output: [] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/substring-with-concatenation-of-all-words/ -// discuss: https://leetcode.com/problems/substring-with-concatenation-of-all-words/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -struct Term { - expect: i32, - count: i32, -} -impl Term { - fn new(expect: i32, count: i32) -> Self { - Term { expect, count } - } - fn inc_expect(&mut self) { - self.expect += 1; - } - fn inc(&mut self) { - self.count += 1; - } - fn dec(&mut self) { - self.count -= 1; - } - fn exhausted(&self) -> bool { - self.count > self.expect - } - fn reset(&mut self) { - self.count = 0; - } -} - -use std::collections::hash_map::Entry; -use std::collections::HashMap; - -impl Solution { - pub fn find_substring(s: String, words: Vec) -> Vec { - if words.len() < 1 { - return vec![]; - } - let word_len = words[0].len(); - if word_len < 1 { - return vec![]; - } - let substr_len = word_len * words.len(); - let mut map: HashMap<&str, Term> = HashMap::with_capacity(words.len()); - for word in words.iter() { - map.entry(word).or_insert(Term::new(0, 0)).inc_expect(); - } - let mut result: Vec = Vec::new(); - // we can split terms in N ways, where N = word_len - for shift in 0..word_len { - let mut i = shift; - let mut j = shift; - // we do a sliding window for each round - while j + word_len - 1 < s.len() { - match map.entry(&s[j..j + word_len]) { - Entry::Occupied(mut entry) => { - entry.get_mut().inc(); - // term exhausted, shrink the window to release - if entry.get().exhausted() { - while i < j { - let term = &s[i..i + word_len]; - map.entry(term).and_modify(|t| t.dec()); - i += word_len; - if term == &s[j..j + word_len] { - break; - } - } - j += word_len; - } else { - if j - i < (words.len() - 1) * word_len { - j += word_len; - } else { - // matched! - result.push(i as i32); - // move the whole window, release the dropped term - map.entry(&s[i..i + word_len]).and_modify(|t| t.dec()); - j += word_len; - i += word_len; - } - } - } - // bad term, move over and do a reset - Entry::Vacant(entry) => { - map.iter_mut().for_each(|(_, v)| v.reset()); - j += word_len; - i = j; - } - } - } - map.iter_mut().for_each(|(_, v)| v.reset()) - } - result - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_30() { - assert_eq!( - Solution::find_substring( - "barfoothefoobarman".to_string(), - vec!["foo".to_string(), "bar".to_string()] - ), - vec![0, 9] - ); - assert_eq!( - Solution::find_substring( - "wordgoodgoodgoodbestword".to_string(), - vec![ - "word".to_string(), - "good".to_string(), - "best".to_string(), - "word".to_string() - ] - ), - vec![] - ); - assert_eq!( - Solution::find_substring( - "wordgoodgoodgoodbestword".to_string(), - vec![ - "word".to_string(), - "good".to_string(), - "best".to_string(), - "good".to_string() - ] - ), - vec![8] - ); - assert_eq!( - Solution::find_substring( - "xxwordgoodgoodgoodbestword".to_string(), - vec![ - "word".to_string(), - "good".to_string(), - "best".to_string(), - "good".to_string() - ] - ), - vec![10] - ); - } -} diff --git a/src/solution/s0031_next_permutation.rs b/src/solution/s0031_next_permutation.rs deleted file mode 100644 index b448dfa1..00000000 --- a/src/solution/s0031_next_permutation.rs +++ /dev/null @@ -1,89 +0,0 @@ -/** - * [31] Next Permutation - * - * Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. - * - * If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). - * - * The replacement must be in-place and use only constant extra memory. - * - * Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. - * - * 1,2,3 → 1,3,2
- * 3,2,1 → 1,2,3
- * 1,1,5 → 1,5,1 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/next-permutation/ -// discuss: https://leetcode.com/problems/next-permutation/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn next_permutation(nums: &mut Vec) { - let len = nums.len(); - let mut i = (len - 1) as i32; - let mut prev = -1; - // find the decrement digit from end - while i >= 0 { - if nums[i as usize] < prev { - break; - } - prev = nums[i as usize]; - i -= 1; - } - let mut j = len - 1; - // find the first digit larger than nums[i] - // we can do binary search here to make a slightly improvement - if i >= 0 { - while j > (i as usize) { - if nums[j] > nums[i as usize] { - nums.swap(i as usize, j); - break; - } - j -= 1; - } - } - let slice = &mut nums[((i + 1) as usize)..len]; - slice.reverse(); - } -} - -// submission codes end - -/* -// a clean solution (from leetcode submissions) -impl Solution { - pub fn next_permutation(a: &mut Vec) { - let n = a.len(); - - if let Some(i) = (1..n).rev().find(|&i| a[i - 1] < a[i]) { - let j = (i..n).rev().find(|&j| a[i - 1] < a[j]) - .unwrap(); - - a.swap(i - 1, j); - a[i..].reverse(); - } else { - a.reverse(); - } - } -} - */ - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_31() { - let mut vec1 = vec![1, 2, 3, 4, 5]; - Solution::next_permutation(&mut vec1); - assert_eq!(vec1, vec![1, 2, 3, 5, 4]); - - let mut vec2 = vec![5, 4, 3, 2, 1]; - Solution::next_permutation(&mut vec2); - assert_eq!(vec2, vec![1, 2, 3, 4, 5]); - } -} diff --git a/src/solution/s0032_longest_valid_parentheses.rs b/src/solution/s0032_longest_valid_parentheses.rs deleted file mode 100644 index d970de2b..00000000 --- a/src/solution/s0032_longest_valid_parentheses.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [32] Longest Valid Parentheses - * - * Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. - * - * Example 1: - * - * - * Input: "(()" - * Output: 2 - * Explanation: The longest valid parentheses substring is "()" - * - * - * Example 2: - * - * - * Input: ")()())" - * Output: 4 - * Explanation: The longest valid parentheses substring is "()()" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-valid-parentheses/ -// discuss: https://leetcode.com/problems/longest-valid-parentheses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// time: O(N) space: O(1) -impl Solution { - pub fn longest_valid_parentheses(s: String) -> i32 { - let mut seq: Vec = s.chars().collect(); - let forward_max = Solution::longest(&seq, '('); - seq.reverse(); - let backward_max = Solution::longest(&seq, ')'); - i32::max(forward_max, backward_max) - } - - fn longest(seq: &Vec, plus_char: char) -> i32 { - let mut stack = 0; - let mut max_len = 0; - let (mut i, mut j) = (0_usize, 0_usize); - while j < seq.len() { - if seq[j] == plus_char { - stack += 1; - } else { - // stack exhausted, shift over - if stack < 1 { - i = j + 1; - } else { - stack -= 1; - if stack == 0 { - max_len = i32::max(max_len, (j - i + 1) as i32); - } - } - } - j += 1; - } - max_len - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_32() { - assert_eq!(Solution::longest_valid_parentheses(")()())".to_string()), 4); - assert_eq!(Solution::longest_valid_parentheses(")(".to_string()), 0); - assert_eq!(Solution::longest_valid_parentheses("(()".to_string()), 2); - assert_eq!( - Solution::longest_valid_parentheses("(((((()()".to_string()), - 4 - ); - assert_eq!( - Solution::longest_valid_parentheses("((((((((()))".to_string()), - 6 - ); - assert_eq!(Solution::longest_valid_parentheses("()".to_string()), 2); - assert_eq!(Solution::longest_valid_parentheses("()(()".to_string()), 2); - assert_eq!( - Solution::longest_valid_parentheses(")()(((())))(".to_string()), - 10 - ); - assert_eq!( - Solution::longest_valid_parentheses("(()(((()".to_string()), - 2 - ); - assert_eq!(Solution::longest_valid_parentheses("".to_string()), 0); - } -} diff --git a/src/solution/s0033_search_in_rotated_sorted_array.rs b/src/solution/s0033_search_in_rotated_sorted_array.rs deleted file mode 100644 index ab5c13d3..00000000 --- a/src/solution/s0033_search_in_rotated_sorted_array.rs +++ /dev/null @@ -1,105 +0,0 @@ -/** - * [33] Search in Rotated Sorted Array - * - * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. - * - * (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). - * - * You are given a target value to search. If found in the array return its index, otherwise return -1. - * - * You may assume no duplicate exists in the array. - * - * Your algorithm's runtime complexity must be in the order of O(log n). - * - * Example 1: - * - * - * Input: nums = [4,5,6,7,0,1,2], target = 0 - * Output: 4 - * - * - * Example 2: - * - * - * Input: nums = [4,5,6,7,0,1,2], target = 3 - * Output: -1 - * - */ -pub struct Solution {} - -/* - \ - 8 - 7 9 - 6 1 - 5 2 - 3 \ -Consider the given array as ring, each time we split the ring and judge which part is the target belong to, then it's ordinary binary search. -*/ - -// problem: https://leetcode.com/problems/search-in-rotated-sorted-array/ -// discuss: https://leetcode.com/problems/search-in-rotated-sorted-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn search(nums: Vec, target: i32) -> i32 { - let mut size = nums.len(); - if size == 0 { - return -1; - } - let mut base = 0_usize; - while size > 1 { - let half = size / 2; - let mid = base + half; - if nums[mid] == target { - return mid as i32; - } - // we split the ring to [base..half] & [half+1..base-1] - // if target not in [base..half] ring, move base to select another ring - if !(((nums[base] < nums[mid]) && (target >= nums[base] && target <= nums[mid])) - || ((nums[base] > nums[mid]) && (target >= nums[base] || target <= nums[mid]))) - { - base = mid; - } - size -= half; - } - if nums[base] == target { - base as i32 - } else { - -1 - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_33() { - assert_eq!(Solution::search(vec![7, 8, 1, 2, 3, 4, 5, 6], 2), 3); - assert_eq!( - Solution::search( - vec![ - 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011, 1012, 0, 1, 2, 3, 4, 5, 6, 7, 8 - ], - 0 - ), - 9 - ); - assert_eq!( - Solution::search( - vec![ - 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011, 1012, 0, 1, 2, 3, 4, 5, 6, 7, 8 - ], - 1006 - ), - 2 - ); - assert_eq!(Solution::search(vec![4, 5, 6, 7, 0, 1, 2], 3), -1); - assert_eq!(Solution::search(vec![], 3), -1); - } -} diff --git a/src/solution/s0034_find_first_and_last_position_of_element_in_sorted_array.rs b/src/solution/s0034_find_first_and_last_position_of_element_in_sorted_array.rs deleted file mode 100644 index eba1c01c..00000000 --- a/src/solution/s0034_find_first_and_last_position_of_element_in_sorted_array.rs +++ /dev/null @@ -1,45 +0,0 @@ -/** - * [34] Find First and Last Position of Element in Sorted Array - * - * Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. - * - * Your algorithm's runtime complexity must be in the order of O(log n). - * - * If the target is not found in the array, return [-1, -1]. - * - * Example 1: - * - * - * Input: nums = [5,7,7,8,8,10], target = 8 - * Output: [3,4] - * - * Example 2: - * - * - * Input: nums = [5,7,7,8,8,10], target = 6 - * Output: [-1,-1] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ -// discuss: https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -impl Solution { - pub fn search_range(nums: Vec, target: i32) -> Vec { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_34() {} -} diff --git a/src/solution/s0035_search_insert_position.rs b/src/solution/s0035_search_insert_position.rs deleted file mode 100644 index 31c1c38b..00000000 --- a/src/solution/s0035_search_insert_position.rs +++ /dev/null @@ -1,59 +0,0 @@ -/** - * [35] Search Insert Position - * - * Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. - * - * You may assume no duplicates in the array. - * - * Example 1: - * - * - * Input: [1,3,5,6], 5 - * Output: 2 - * - * - * Example 2: - * - * - * Input: [1,3,5,6], 2 - * Output: 1 - * - * - * Example 3: - * - * - * Input: [1,3,5,6], 7 - * Output: 4 - * - * - * Example 4: - * - * - * Input: [1,3,5,6], 0 - * Output: 0 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/search-insert-position/ -// discuss: https://leetcode.com/problems/search-insert-position/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -impl Solution { - pub fn search_insert(nums: Vec, target: i32) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_35() {} -} diff --git a/src/solution/s0036_valid_sudoku.rs b/src/solution/s0036_valid_sudoku.rs deleted file mode 100644 index 5a595774..00000000 --- a/src/solution/s0036_valid_sudoku.rs +++ /dev/null @@ -1,172 +0,0 @@ -/** - * [36] Valid Sudoku - * - * Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules: - * - *
    - * Each row must contain the digits 1-9 without repetition. - * Each column must contain the digits 1-9 without repetition. - * Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition. - *
- * - *
- * A partially filled sudoku which is valid. - * - * The Sudoku board could be partially filled, where empty cells are filled with the character '.'. - * - * Example 1: - * - * - * Input: - * [ - * ["5","3",".",".","7",".",".",".","."], - * ["6",".",".","1","9","5",".",".","."], - * [".","9","8",".",".",".",".","6","."], - * ["8",".",".",".","6",".",".",".","3"], - * ["4",".",".","8",".","3",".",".","1"], - * ["7",".",".",".","2",".",".",".","6"], - * [".","6",".",".",".",".","2","8","."], - * [".",".",".","4","1","9",".",".","5"], - * [".",".",".",".","8",".",".","7","9"] - * ] - * Output: true - * - * - * Example 2: - * - * - * Input: - * [ - * ["8","3",".",".","7",".",".",".","."], - * ["6",".",".","1","9","5",".",".","."], - * [".","9","8",".",".",".",".","6","."], - * ["8",".",".",".","6",".",".",".","3"], - * ["4",".",".","8",".","3",".",".","1"], - * ["7",".",".",".","2",".",".",".","6"], - * [".","6",".",".",".",".","2","8","."], - * [".",".",".","4","1","9",".",".","5"], - * [".",".",".",".","8",".",".","7","9"] - * ] - * Output: false - * Explanation: Same as Example 1, except with the 5 in the top left corner being - * modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid. - * - * - * Note: - * - * - * A Sudoku board (partially filled) could be valid but is not necessarily solvable. - * Only the filled cells need to be validated according to the mentioned rules. - * The given board contain only digits 1-9 and the character '.'. - * The given board size is always 9x9. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/valid-sudoku/ -// discuss: https://leetcode.com/problems/valid-sudoku/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// just brute force -impl Solution { - pub fn is_valid_sudoku(board: Vec>) -> bool { - let mut table = vec![0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; - for row in board.iter() { - for z in 1..10 { - table[z] = 0; - } - for ch in row { - match ch.to_digit(10) { - None => continue, - Some(idx) => { - if table[idx as usize] > 0 { - return false; - } else { - table[idx as usize] = 1 - } - } - } - } - } - for i in 0..9 { - for z in 1..10 { - table[z] = 0; - } - for row in board.iter() { - match row[i].to_digit(10) { - None => continue, - Some(idx) => { - if table[idx as usize] > 0 { - return false; - } else { - table[idx as usize] = 1 - } - } - } - } - } - for i in 0..3 { - for j in 0..3 { - for z in 1..10 { - table[z] = 0; - } - for row in 3 * i..3 * (i + 1) { - for column in 3 * j..3 * (j + 1) { - match board[row][column].to_digit(10) { - None => continue, - Some(idx) => { - if table[idx as usize] > 0 { - return false; - } else { - table[idx as usize] = 1 - } - } - } - } - } - } - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_36() { - assert_eq!( - Solution::is_valid_sudoku(vec![ - vec!['8', '3', '.', '.', '7', '.', '.', '.', '.'], - vec!['6', '.', '.', '1', '9', '5', '.', '.', '.'], - vec!['.', '9', '8', '.', '.', '.', '.', '6', '.'], - vec!['8', '.', '.', '.', '6', '.', '.', '.', '3'], - vec!['4', '.', '.', '8', '.', '3', '.', '.', '1'], - vec!['7', '.', '.', '.', '2', '.', '.', '.', '6'], - vec!['.', '6', '.', '.', '.', '.', '2', '8', '.'], - vec!['.', '.', '.', '4', '1', '9', '.', '.', '5'], - vec!['.', '.', '.', '.', '8', '.', '.', '7', '9'], - ]), - false - ); - assert_eq!( - Solution::is_valid_sudoku(vec![ - vec!['5', '3', '.', '.', '7', '.', '.', '.', '.'], - vec!['6', '.', '.', '1', '9', '5', '.', '.', '.'], - vec!['.', '9', '8', '.', '.', '.', '.', '6', '.'], - vec!['8', '.', '.', '.', '6', '.', '.', '.', '3'], - vec!['4', '.', '.', '8', '.', '3', '.', '.', '1'], - vec!['7', '.', '.', '.', '2', '.', '.', '.', '6'], - vec!['.', '6', '.', '.', '.', '.', '2', '8', '.'], - vec!['.', '.', '.', '4', '1', '9', '.', '.', '5'], - vec!['.', '.', '.', '.', '8', '.', '.', '7', '9'] - ]), - true - ); - } -} diff --git a/src/solution/s0037_sudoku_solver.rs b/src/solution/s0037_sudoku_solver.rs deleted file mode 100644 index d0187582..00000000 --- a/src/solution/s0037_sudoku_solver.rs +++ /dev/null @@ -1,51 +0,0 @@ -/** - * [37] Sudoku Solver - * - * Write a program to solve a Sudoku puzzle by filling the empty cells. - * - * A sudoku solution must satisfy all of the following rules: - * - *
    - * Each of the digits 1-9 must occur exactly once in each row. - * Each of the digits 1-9 must occur exactly once in each column. - * Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid. - *
- * - * Empty cells are indicated by the character '.'. - * - *
- * A sudoku puzzle... - * - *
- * ...and its solution numbers marked in red. - * - * Note: - * - * - * The given board contain only digits 1-9 and the character '.'. - * You may assume that the given Sudoku puzzle will have a single unique solution. - * The given board size is always 9x9. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/sudoku-solver/ -// discuss: https://leetcode.com/problems/sudoku-solver/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -impl Solution { - pub fn solve_sudoku(board: &mut Vec>) {} -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_37() {} -} diff --git a/src/solution/s0038_count_and_say.rs b/src/solution/s0038_count_and_say.rs deleted file mode 100644 index 6305b2a5..00000000 --- a/src/solution/s0038_count_and_say.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [38] Count and Say - * - * The count-and-say sequence is the sequence of integers with the first five terms as following: - * - * - * 1. 1 - * 2. 11 - * 3. 21 - * 4. 1211 - * 5. 111221 - * - * - * 1 is read off as "one 1" or 11.
- * 11 is read off as "two 1s" or 21.
- * 21 is read off as "one 2, then one 1" or 1211. - * - * Given an integer n where 1 <= n <= 30, generate the n^th term of the count-and-say sequence. - * - * Note: Each term of the sequence of integers will be represented as a string. - * - * - * - * Example 1: - * - * - * Input: 1 - * Output: "1" - * - * - * Example 2: - * - * - * Input: 4 - * Output: "1211" - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/count-and-say/ -// discuss: https://leetcode.com/problems/count-and-say/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::char::from_digit; -impl Solution { - pub fn count_and_say(n: i32) -> String { - let mut res = vec!['1']; - for _ in 0..n - 1 { - let mut temp = Vec::new(); - let mut i = 0_usize; - while i < res.len() { - let mut j = i + 1; - while j < res.len() && res[j] == res[i] { - j += 1; - } - temp.push(from_digit((j - i) as u32, 10).unwrap()); - temp.push(res[i]); - i = j; - } - res = temp; - } - res.iter().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_38() { - assert_eq!(Solution::count_and_say(1), "1"); - assert_eq!(Solution::count_and_say(2), "11"); - assert_eq!(Solution::count_and_say(3), "21"); - assert_eq!(Solution::count_and_say(4), "1211"); - assert_eq!(Solution::count_and_say(5), "111221"); - } -} diff --git a/src/solution/s0039_combination_sum.rs b/src/solution/s0039_combination_sum.rs deleted file mode 100644 index b31e397e..00000000 --- a/src/solution/s0039_combination_sum.rs +++ /dev/null @@ -1,100 +0,0 @@ -/** - * [39] Combination Sum - * - * Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. - * - * The same repeated number may be chosen from candidates unlimited number of times. - * - * Note: - * - * - * All numbers (including target) will be positive integers. - * The solution set must not contain duplicate combinations. - * - * - * Example 1: - * - * - * Input: candidates = [2,3,6,7], target = 7, - * A solution set is: - * [ - * [7], - * [2,2,3] - * ] - * - * - * Example 2: - * - * - * Input: candidates = [2,3,5], target = 8, - * A solution set is: - * [ - * [2,2,2,2], - * [2,3,3], - * [3,5] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/combination-sum/ -// discuss: https://leetcode.com/problems/combination-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn combination_sum(candidates: Vec, target: i32) -> Vec> { - let mut seq = candidates; - let mut res = Vec::new(); - seq.sort_unstable_by(|a, b| b.cmp(a)); - let mut vec = Vec::new(); - Solution::backtrack(&seq, target, vec, &mut res, 0); - res - } - - fn backtrack( - seq: &Vec, - target: i32, - mut curr: Vec, - result: &mut Vec>, - start_idx: usize, - ) { - for i in start_idx..seq.len() { - let item = seq[i]; - if target - item < 0 { - continue; - } - let mut new_vec = curr.clone(); - new_vec.push(item); - if target == item { - result.push(new_vec); - } else { - Solution::backtrack(seq, target - item, new_vec, result, i); - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_39() { - assert_eq!( - Solution::combination_sum(vec![1], 7), - vec![vec![1, 1, 1, 1, 1, 1, 1]] - ); - assert_eq!( - Solution::combination_sum(vec![2, 3, 6, 7], 7), - vec![vec![7], vec![3, 2, 2],] - ); - assert_eq!( - Solution::combination_sum(vec![2, 3, 5], 8), - vec![vec![5, 3], vec![3, 3, 2], vec![2, 2, 2, 2],] - ); - } -} diff --git a/src/solution/s0040_combination_sum_ii.rs b/src/solution/s0040_combination_sum_ii.rs deleted file mode 100644 index 0445bfb9..00000000 --- a/src/solution/s0040_combination_sum_ii.rs +++ /dev/null @@ -1,107 +0,0 @@ -/** - * [40] Combination Sum II - * - * Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. - * - * Each number in candidates may only be used once in the combination. - * - * Note: - * - * - * All numbers (including target) will be positive integers. - * The solution set must not contain duplicate combinations. - * - * - * Example 1: - * - * - * Input: candidates = [10,1,2,7,6,1,5], target = 8, - * A solution set is: - * [ - * [1, 7], - * [1, 2, 5], - * [2, 6], - * [1, 1, 6] - * ] - * - * - * Example 2: - * - * - * Input: candidates = [2,5,2,1,2], target = 5, - * A solution set is: - * [ - * [1,2,2], - * [5] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/combination-sum-ii/ -// discuss: https://leetcode.com/problems/combination-sum-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn combination_sum2(candidates: Vec, target: i32) -> Vec> { - let mut seq = candidates; - let mut res = Vec::new(); - seq.sort_unstable_by(|a, b| b.cmp(a)); - let mut vec = Vec::new(); - Solution::backtrack(&seq, target, vec, &mut res, 0); - res - } - - fn backtrack( - seq: &Vec, - target: i32, - mut curr: Vec, - result: &mut Vec>, - start_idx: usize, - ) { - let mut i = start_idx; - while i < seq.len() { - let item = seq[i]; - if target - item < 0 { - i += 1; - continue; - } - let mut new_vec = curr.clone(); - new_vec.push(item); - if target == item { - result.push(new_vec); - } else { - Solution::backtrack(seq, target - item, new_vec, result, i + 1); - } - // skip duplicate result - while i < seq.len() && seq[i] == item { - i += 1; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_40() { - assert_eq!( - Solution::combination_sum2(vec![1, 1, 1, 1, 1, 1, 1], 7), - vec![vec![1, 1, 1, 1, 1, 1, 1]] - ); - assert_eq!( - Solution::combination_sum2(vec![10, 1, 2, 7, 6, 1, 5], 8), - vec![vec![7, 1], vec![6, 2], vec![6, 1, 1], vec![5, 2, 1],] - ); - assert_eq!( - Solution::combination_sum2(vec![2, 5, 2, 1, 2], 5), - vec![vec![5], vec![2, 2, 1],] - ); - } -} diff --git a/src/solution/s0041_first_missing_positive.rs b/src/solution/s0041_first_missing_positive.rs deleted file mode 100644 index 80cd8d8e..00000000 --- a/src/solution/s0041_first_missing_positive.rs +++ /dev/null @@ -1,91 +0,0 @@ -/** - * [41] First Missing Positive - * - * Given an unsorted integer array, find the smallest missing positive integer. - * - * Example 1: - * - * - * Input: [1,2,0] - * Output: 3 - * - * - * Example 2: - * - * - * Input: [3,4,-1,1] - * Output: 2 - * - * - * Example 3: - * - * - * Input: [7,8,9,11,12] - * Output: 1 - * - * - * Note: - * - * Your algorithm should run in O(n) time and uses constant extra space. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/first-missing-positive/ -// discuss: https://leetcode.com/problems/first-missing-positive/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn first_missing_positive(nums: Vec) -> i32 { - let mut nums = nums; - let len = nums.len(); - let mut i = 0; - let mut c = 0; - while i < len { - let num = nums[i]; - if num > 0 && num - 1 < (len as i32) { - c += 1; - nums.swap((num - 1) as usize, i); - if (num - 1) > (i as i32) && (num != nums[i]) { - continue; - } - } - i += 1; - } - println!("{}", c); - for (i, &num) in nums.iter().enumerate() { - if num != ((i + 1) as i32) { - return (i + 1) as i32; - } - } - return (len + 1) as i32; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_41() { - assert_eq!(Solution::first_missing_positive(vec![2, 2]), 1); - assert_eq!( - Solution::first_missing_positive(vec![12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2]), - 1 - ); - assert_eq!( - Solution::first_missing_positive(vec![2, 2, 2, 2, 2, 2, 2]), - 1 - ); - assert_eq!(Solution::first_missing_positive(vec![3, 4, -1, 1]), 2); - assert_eq!(Solution::first_missing_positive(vec![2, 1, 0]), 3); - assert_eq!(Solution::first_missing_positive(vec![7, 8, 9, 11, 12]), 1); - assert_eq!( - Solution::first_missing_positive(vec![7, 8, 1, 2, 3, 3, 3, 3, 3, 3, 3, -5, -7, 1234]), - 4 - ); - } -} diff --git a/src/solution/s0042_trapping_rain_water.rs b/src/solution/s0042_trapping_rain_water.rs deleted file mode 100644 index 9d961994..00000000 --- a/src/solution/s0042_trapping_rain_water.rs +++ /dev/null @@ -1,38 +0,0 @@ -/** - * [42] Trapping Rain Water - * - * Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. - * - *
- * The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image! - * - * Example: - * - * - * Input: [0,1,0,2,1,0,1,3,2,1,2,1] - * Output: 6 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/trapping-rain-water/ -// discuss: https://leetcode.com/problems/trapping-rain-water/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -impl Solution { - pub fn trap(height: Vec) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_42() {} -} diff --git a/src/solution/s0043_multiply_strings.rs b/src/solution/s0043_multiply_strings.rs deleted file mode 100644 index 8c487b8e..00000000 --- a/src/solution/s0043_multiply_strings.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [43] Multiply Strings - * - * Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. - * - * Example 1: - * - * - * Input: num1 = "2", num2 = "3" - * Output: "6" - * - * Example 2: - * - * - * Input: num1 = "123", num2 = "456" - * Output: "56088" - * - * - * Note: - * - *
    - * The length of both num1 and num2 is < 110. - * Both num1 and num2 contain only digits 0-9. - * Both num1 and num2 do not contain any leading zero, except the number 0 itself. - * You must not use any built-in BigInteger library or convert the inputs to integer directly. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/multiply-strings/ -// discuss: https://leetcode.com/problems/multiply-strings/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -use std::char::from_digit; -use std::collections::VecDeque; -impl Solution { - pub fn multiply(num1: String, num2: String) -> String { - let mut num1: Vec = num1.chars().map(|ch| ch.to_digit(10).unwrap()).collect(); - let mut num2: Vec = num2.chars().map(|ch| ch.to_digit(10).unwrap()).collect(); - let mut buffer = VecDeque::with_capacity(num2.len() + 1); - let mut res: Vec = Vec::new(); - let mut carry = 0_u32; - num1.reverse(); - num2.reverse(); - for (i, multiplier) in num1.into_iter().enumerate() { - buffer - .pop_back() - .and_then(|digit| Some(res.push(from_digit(digit, 10).unwrap()))); - for &multiplicand in num2.iter() {} - } - res.reverse(); - res.into_iter().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_43() {} -} diff --git a/src/solution/s0044_wildcard_matching.rs b/src/solution/s0044_wildcard_matching.rs deleted file mode 100644 index e875131a..00000000 --- a/src/solution/s0044_wildcard_matching.rs +++ /dev/null @@ -1,91 +0,0 @@ -/** - * [44] Wildcard Matching - * - * Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'. - * - * - * '?' Matches any single character. - * '*' Matches any sequence of characters (including the empty sequence). - * - * - * The matching should cover the entire input string (not partial). - * - * Note: - * - * - * s could be empty and contains only lowercase letters a-z. - * p could be empty and contains only lowercase letters a-z, and characters like ? or *. - * - * - * Example 1: - * - * - * Input: - * s = "aa" - * p = "a" - * Output: false - * Explanation: "a" does not match the entire string "aa". - * - * - * Example 2: - * - * - * Input: - * s = "aa" - * p = "*" - * Output: true - * Explanation: '*' matches any sequence. - * - * - * Example 3: - * - * - * Input: - * s = "cb" - * p = "?a" - * Output: false - * Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'. - * - * - * Example 4: - * - * - * Input: - * s = "adceb" - * p = "*a*b" - * Output: true - * Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce". - * - * - * Example 5: - * - * - * Input: - * s = "acdcb" - * p = "a*c?b" - * Output: false - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/wildcard-matching/ -// discuss: https://leetcode.com/problems/wildcard-matching/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_match(s: String, p: String) -> bool { - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_44() {} -} diff --git a/src/solution/s0045_jump_game_ii.rs b/src/solution/s0045_jump_game_ii.rs deleted file mode 100644 index 3922e20b..00000000 --- a/src/solution/s0045_jump_game_ii.rs +++ /dev/null @@ -1,45 +0,0 @@ -/** - * [45] Jump Game II - * - * Given an array of non-negative integers, you are initially positioned at the first index of the array. - * - * Each element in the array represents your maximum jump length at that position. - * - * Your goal is to reach the last index in the minimum number of jumps. - * - * Example: - * - * - * Input: [2,3,1,1,4] - * Output: 2 - * Explanation: The minimum number of jumps to reach the last index is 2. - * Jump 1 step from index 0 to 1, then 3 steps to the last index. - * - * Note: - * - * You can assume that you can always reach the last index. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/jump-game-ii/ -// discuss: https://leetcode.com/problems/jump-game-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: shortest path from backward -impl Solution { - pub fn jump(nums: Vec) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_45() {} -} diff --git a/src/solution/s0046_permutations.rs b/src/solution/s0046_permutations.rs deleted file mode 100644 index ef9354c8..00000000 --- a/src/solution/s0046_permutations.rs +++ /dev/null @@ -1,70 +0,0 @@ -/** - * [46] Permutations - * - * Given a collection of distinct integers, return all possible permutations. - * - * Example: - * - * - * Input: [1,2,3] - * Output: - * [ - * [1,2,3], - * [1,3,2], - * [2,1,3], - * [2,3,1], - * [3,1,2], - * [3,2,1] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/permutations/ -// discuss: https://leetcode.com/problems/permutations/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn permute(nums: Vec) -> Vec> { - if nums.len() <= 1 { - return vec![nums]; - } - nums.iter() - .flat_map(|&num| { - let mut sub = nums.clone().into_iter().filter(|&x| x != num).collect(); - Solution::permute(sub) - .into_iter() - .map(|vec| { - let mut vec = vec; - vec.push(num); - vec - }) - .collect::>>() - }) - .collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_46() { - assert_eq!( - Solution::permute(vec![1, 2, 3]), - vec![ - vec![3, 2, 1], - vec![2, 3, 1], - vec![3, 1, 2], - vec![1, 3, 2], - vec![2, 1, 3], - vec![1, 2, 3], - ] - ) - } -} diff --git a/src/solution/s0047_permutations_ii.rs b/src/solution/s0047_permutations_ii.rs deleted file mode 100644 index 7dceb55b..00000000 --- a/src/solution/s0047_permutations_ii.rs +++ /dev/null @@ -1,179 +0,0 @@ -/** - * [47] Permutations II - * - * Given a collection of numbers that might contain duplicates, return all possible unique permutations. - * - * Example: - * - * - * Input: [1,1,2] - * Output: - * [ - * [1,1,2], - * [1,2,1], - * [2,1,1] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/permutations-ii/ -// discuss: https://leetcode.com/problems/permutations-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn permute_unique(nums: Vec) -> Vec> { - let mut nums = nums; - nums.sort_unstable(); - Solution::permute(nums) - } - - fn permute(mut nums: Vec) -> Vec> { - if nums.len() <= 1 { - return vec![nums]; - } - let mut prev: Option = None; - let mut res = Vec::new(); - for (i, &num) in nums.iter().enumerate() { - if prev.is_some() && prev.unwrap() == num { - continue; - } else { - prev = Some(num) - } - let mut sub = nums.clone(); - sub.remove(i); - let mut permutations: Vec> = Solution::permute(sub) - .into_iter() - .map(|x| { - let mut x = x; - x.push(num); - x - }) - .collect(); - res.append(&mut permutations); - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_47() { - assert_eq!( - Solution::permute(vec![1, 1, 2]), - vec![vec![2, 1, 1], vec![1, 2, 1], vec![1, 1, 2],] - ); - assert_eq!(Solution::permute(vec![1, 1, 1]), vec![vec![1, 1, 1],]); - assert_eq!( - Solution::permute(vec![1, 1, 1, 2]), - vec![ - vec![2, 1, 1, 1], - vec![1, 2, 1, 1], - vec![1, 1, 2, 1], - vec![1, 1, 1, 2], - ] - ); - assert_eq!( - Solution::permute(vec![1, 1, 2, 2, 3, 3]), - vec![ - vec![3, 3, 2, 2, 1, 1], - vec![3, 2, 3, 2, 1, 1], - vec![2, 3, 3, 2, 1, 1], - vec![3, 2, 2, 3, 1, 1], - vec![2, 3, 2, 3, 1, 1], - vec![2, 2, 3, 3, 1, 1], - vec![3, 3, 2, 1, 2, 1], - vec![3, 2, 3, 1, 2, 1], - vec![2, 3, 3, 1, 2, 1], - vec![3, 3, 1, 2, 2, 1], - vec![3, 1, 3, 2, 2, 1], - vec![1, 3, 3, 2, 2, 1], - vec![3, 2, 1, 3, 2, 1], - vec![2, 3, 1, 3, 2, 1], - vec![3, 1, 2, 3, 2, 1], - vec![1, 3, 2, 3, 2, 1], - vec![2, 1, 3, 3, 2, 1], - vec![1, 2, 3, 3, 2, 1], - vec![3, 2, 2, 1, 3, 1], - vec![2, 3, 2, 1, 3, 1], - vec![2, 2, 3, 1, 3, 1], - vec![3, 2, 1, 2, 3, 1], - vec![2, 3, 1, 2, 3, 1], - vec![3, 1, 2, 2, 3, 1], - vec![1, 3, 2, 2, 3, 1], - vec![2, 1, 3, 2, 3, 1], - vec![1, 2, 3, 2, 3, 1], - vec![2, 2, 1, 3, 3, 1], - vec![2, 1, 2, 3, 3, 1], - vec![1, 2, 2, 3, 3, 1], - vec![3, 3, 2, 1, 1, 2], - vec![3, 2, 3, 1, 1, 2], - vec![2, 3, 3, 1, 1, 2], - vec![3, 3, 1, 2, 1, 2], - vec![3, 1, 3, 2, 1, 2], - vec![1, 3, 3, 2, 1, 2], - vec![3, 2, 1, 3, 1, 2], - vec![2, 3, 1, 3, 1, 2], - vec![3, 1, 2, 3, 1, 2], - vec![1, 3, 2, 3, 1, 2], - vec![2, 1, 3, 3, 1, 2], - vec![1, 2, 3, 3, 1, 2], - vec![3, 3, 1, 1, 2, 2], - vec![3, 1, 3, 1, 2, 2], - vec![1, 3, 3, 1, 2, 2], - vec![3, 1, 1, 3, 2, 2], - vec![1, 3, 1, 3, 2, 2], - vec![1, 1, 3, 3, 2, 2], - vec![3, 2, 1, 1, 3, 2], - vec![2, 3, 1, 1, 3, 2], - vec![3, 1, 2, 1, 3, 2], - vec![1, 3, 2, 1, 3, 2], - vec![2, 1, 3, 1, 3, 2], - vec![1, 2, 3, 1, 3, 2], - vec![3, 1, 1, 2, 3, 2], - vec![1, 3, 1, 2, 3, 2], - vec![1, 1, 3, 2, 3, 2], - vec![2, 1, 1, 3, 3, 2], - vec![1, 2, 1, 3, 3, 2], - vec![1, 1, 2, 3, 3, 2], - vec![3, 2, 2, 1, 1, 3], - vec![2, 3, 2, 1, 1, 3], - vec![2, 2, 3, 1, 1, 3], - vec![3, 2, 1, 2, 1, 3], - vec![2, 3, 1, 2, 1, 3], - vec![3, 1, 2, 2, 1, 3], - vec![1, 3, 2, 2, 1, 3], - vec![2, 1, 3, 2, 1, 3], - vec![1, 2, 3, 2, 1, 3], - vec![2, 2, 1, 3, 1, 3], - vec![2, 1, 2, 3, 1, 3], - vec![1, 2, 2, 3, 1, 3], - vec![3, 2, 1, 1, 2, 3], - vec![2, 3, 1, 1, 2, 3], - vec![3, 1, 2, 1, 2, 3], - vec![1, 3, 2, 1, 2, 3], - vec![2, 1, 3, 1, 2, 3], - vec![1, 2, 3, 1, 2, 3], - vec![3, 1, 1, 2, 2, 3], - vec![1, 3, 1, 2, 2, 3], - vec![1, 1, 3, 2, 2, 3], - vec![2, 1, 1, 3, 2, 3], - vec![1, 2, 1, 3, 2, 3], - vec![1, 1, 2, 3, 2, 3], - vec![2, 2, 1, 1, 3, 3], - vec![2, 1, 2, 1, 3, 3], - vec![1, 2, 2, 1, 3, 3], - vec![2, 1, 1, 2, 3, 3], - vec![1, 2, 1, 2, 3, 3], - vec![1, 1, 2, 2, 3, 3] - ] - ); - } -} diff --git a/src/solution/s0048_rotate_image.rs b/src/solution/s0048_rotate_image.rs deleted file mode 100644 index 6f6fdffa..00000000 --- a/src/solution/s0048_rotate_image.rs +++ /dev/null @@ -1,122 +0,0 @@ -/** - * [48] Rotate Image - * - * You are given an n x n 2D matrix representing an image. - * - * Rotate the image by 90 degrees (clockwise). - * - * Note: - * - * You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation. - * - * Example 1: - * - * - * Given input matrix = - * [ - * [1,2,3], - * [4,5,6], - * [7,8,9] - * ], - * - * rotate the input matrix in-place such that it becomes: - * [ - * [7,4,1], - * [8,5,2], - * [9,6,3] - * ] - * - * - * Example 2: - * - * - * Given input matrix = - * [ - * [ 5, 1, 9,11], - * [ 2, 4, 8,10], - * [13, 3, 6, 7], - * [15,14,12,16] - * ], - * - * rotate the input matrix in-place such that it becomes: - * [ - * [15,13, 2, 5], - * [14, 3, 4, 1], - * [12, 6, 8, 9], - * [16, 7,10,11] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/rotate-image/ -// discuss: https://leetcode.com/problems/rotate-image/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// x,y -> y,n-x 2-dimension vector rotate -90 degree: -// ^ x 0 1 y -// | | * = -// v y -1 0 -x -// n-y,x <- n-x,n-y if we consider axis transform, then: rotate(x, y) = (y, -x + n) - -// we only need to iterate a 1/4 corner matrix, for odd matrix, we take an extra part in x direction -// -// even: -// -// x x o o -// x x o o -// o o o o -// o o o o -// -// odd: -// -// x x o o o -// x x o o o -// x x o o o -// o o o o o -// o o o o o - -impl Solution { - pub fn rotate(matrix: &mut Vec>) { - let mut matrix = matrix; - let (len, n) = (matrix.len(), matrix.len() - 1); - for x in 0..len / 2 { - for y in 0..(len + 1) / 2 { - let temp = matrix[x][y]; - matrix[x][y] = matrix[n - y][x]; - matrix[n - y][x] = matrix[n - x][n - y]; - matrix[n - x][n - y] = matrix[y][n - x]; - matrix[y][n - x] = temp; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_48() { - let mut matrix = vec![ - vec![5, 1, 9, 11], - vec![2, 4, 8, 10], - vec![13, 3, 6, 7], - vec![15, 14, 12, 16], - ]; - Solution::rotate(&mut matrix); - assert_eq!( - matrix, - vec![ - vec![15, 13, 2, 5], - vec![14, 3, 4, 1], - vec![12, 6, 8, 9], - vec![16, 7, 10, 11] - ] - ); - } -} diff --git a/src/solution/s0049_group_anagrams.rs b/src/solution/s0049_group_anagrams.rs deleted file mode 100644 index 64200ea3..00000000 --- a/src/solution/s0049_group_anagrams.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [49] Group Anagrams - * - * Given an array of strings, group anagrams together. - * - * Example: - * - * - * Input: ["eat", "tea", "tan", "ate", "nat", "bat"], - * Output: - * [ - * ["ate","eat","tea"], - * ["nat","tan"], - * ["bat"] - * ] - * - * Note: - * - * - * All inputs will be in lowercase. - * The order of your output does not matter. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/group-anagrams/ -// discuss: https://leetcode.com/problems/group-anagrams/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashMap; -impl Solution { - pub fn group_anagrams(strs: Vec) -> Vec> { - let mut map = HashMap::new(); - for s in strs.into_iter() { - let mut key = [0; 26]; - for ch in s.chars() { - key[(ch as u32 - 'a' as u32) as usize] += 1; - } - map.entry(key).or_insert(Vec::new()).push(s); - } - map.into_iter().map(|(_, v)| v).collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - use std::collections::HashSet; - // TODO: implement arbitrary match macro - #[test] - #[ignore] - fn test_49() { - assert_eq!( - Solution::group_anagrams(vec_string!["eat", "tea", "tan", "ate", "nat", "bat"]), - vec![ - vec_string!["tan", "nat"], - vec_string!["bat"], - vec_string!["eat", "ate", "tea"], - ] - ); - } -} diff --git a/src/solution/s0050_powx_n.rs b/src/solution/s0050_powx_n.rs deleted file mode 100644 index 3e2e9b3e..00000000 --- a/src/solution/s0050_powx_n.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [50] Pow(x, n) - * - * Implement pow(x, n), which calculates x raised to the power n (x^n). - * - * Example 1: - * - * - * Input: 2.00000, 10 - * Output: 1024.00000 - * - * - * Example 2: - * - * - * Input: 2.10000, 3 - * Output: 9.26100 - * - * - * Example 3: - * - * - * Input: 2.00000, -2 - * Output: 0.25000 - * Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25 - * - * - * Note: - * - * - * -100.0 < x < 100.0 - * n is a 32-bit signed integer, within the range [-2^31, 2^31 - 1] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/powx-n/ -// discuss: https://leetcode.com/problems/powx-n/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn my_pow(x: f64, n: i32) -> f64 { - x.powi(n) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_50() { - assert_eq!(Solution::my_pow(2.0, -2), 0.25); - assert_eq!(Solution::my_pow(2.0, 4), 16.0); - assert_eq!(Solution::my_pow(2.0, 5), 32.0); - assert_eq!(Solution::my_pow(2.0, 1), 2.0); - assert_eq!(Solution::my_pow(2.0, -1), 0.5); - assert_eq!(Solution::my_pow(2.0, 10), 1024.0); - } -} diff --git a/src/solution/s0051_n_queens.rs b/src/solution/s0051_n_queens.rs deleted file mode 100644 index d78435a9..00000000 --- a/src/solution/s0051_n_queens.rs +++ /dev/null @@ -1,109 +0,0 @@ -/** - * [51] N-Queens - * - * The n-queens puzzle is the problem of placing n queens on an n*n chessboard such that no two queens attack each other. - * - * - * - * Given an integer n, return all distinct solutions to the n-queens puzzle. - * - * Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively. - * - * Example: - * - * - * Input: 4 - * Output: [ - * [".Q..", // Solution 1 - * "...Q", - * "Q...", - * "..Q."], - * - * ["..Q.", // Solution 2 - * "Q...", - * "...Q", - * ".Q.."] - * ] - * Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/n-queens/ -// discuss: https://leetcode.com/problems/n-queens/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn solve_n_queens(n: i32) -> Vec> { - let mut board = vec![vec!['.'; n as usize]; n as usize]; - let mut solution = Vec::new(); - Solution::schedule_queens(&mut board, &mut solution, n as usize, 0); - solution - } - - fn schedule_queens( - board: &mut Vec>, - solution: &mut Vec>, - len: usize, - row: usize, - ) { - for col in 0..len { - if !Solution::collision(&board, len, row, col) { - board[row][col] = 'Q'; - if row == len - 1 { - solution.push(board.iter().map(|vec| vec.iter().collect()).collect()); - } else { - Solution::schedule_queens(board, solution, len, row + 1); - } - board[row][col] = '.'; - } - } - } - - #[inline(always)] - fn collision(board: &Vec>, len: usize, x: usize, y: usize) -> bool { - for i in 0..x { - if board[i][y] == 'Q' { - return true; - } - } - let (mut i, mut j) = (x as i32 - 1, y as i32 - 1); - while i >= 0 && j >= 0 { - if board[i as usize][j as usize] == 'Q' { - return true; - } - i -= 1; - j -= 1; - } - let (mut i, mut j) = (x as i32 - 1, y as i32 + 1); - while i >= 0 && j < len as i32 { - if board[i as usize][j as usize] == 'Q' { - return true; - } - i -= 1; - j += 1; - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_51() { - assert_eq!( - Solution::solve_n_queens(4), - vec![ - vec![".Q..", "...Q", "Q...", "..Q."], - vec!["..Q.", "Q...", "...Q", ".Q.."] - ] - ); - assert_eq!(Solution::solve_n_queens(8).len(), 92); - } -} diff --git a/src/solution/s0052_n_queens_ii.rs b/src/solution/s0052_n_queens_ii.rs deleted file mode 100644 index 7c9b1987..00000000 --- a/src/solution/s0052_n_queens_ii.rs +++ /dev/null @@ -1,99 +0,0 @@ -/** - * [52] N-Queens II - * - * The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. - * - * - * - * Given an integer n, return the number of distinct solutions to the n-queens puzzle. - * - * Example: - * - * - * Input: 4 - * Output: 2 - * Explanation: There are two distinct solutions to the 4-queens puzzle as shown below. - * [ - * [".Q..", // Solution 1 - * "...Q", - * "Q...", - * "..Q."], - * - * ["..Q.", // Solution 2 - * "Q...", - * "...Q", - * ".Q.."] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/n-queens-ii/ -// discuss: https://leetcode.com/problems/n-queens-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn total_n_queens(n: i32) -> i32 { - let mut board = vec![vec!['.'; n as usize]; n as usize]; - let mut num = 0; - Solution::schedule_queens(&mut board, &mut num, n as usize, 0); - num - } - - fn schedule_queens(board: &mut Vec>, num: &mut i32, len: usize, row: usize) { - for col in 0..len { - if !Solution::collision(&board, len, row, col) { - board[row][col] = 'Q'; - if row == len - 1 { - *num += 1; - } else { - Solution::schedule_queens(board, num, len, row + 1); - } - board[row][col] = '.'; - } - } - } - - #[inline(always)] - fn collision(board: &Vec>, len: usize, x: usize, y: usize) -> bool { - for i in 0..x { - if board[i][y] == 'Q' { - return true; - } - } - let (mut i, mut j) = (x as i32 - 1, y as i32 - 1); - while i >= 0 && j >= 0 { - if board[i as usize][j as usize] == 'Q' { - return true; - } - i -= 1; - j -= 1; - } - let (mut i, mut j) = (x as i32 - 1, y as i32 + 1); - while i >= 0 && j < len as i32 { - if board[i as usize][j as usize] == 'Q' { - return true; - } - i -= 1; - j += 1; - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_52() { - assert_eq!(Solution::total_n_queens(4), 2); - assert_eq!(Solution::total_n_queens(8), 92); - assert_eq!(Solution::total_n_queens(13), 73712); - // assert_eq!(Solution::total_n_queens(14), 365596); - } -} diff --git a/src/solution/s0053_maximum_subarray.rs b/src/solution/s0053_maximum_subarray.rs deleted file mode 100644 index 7967fb00..00000000 --- a/src/solution/s0053_maximum_subarray.rs +++ /dev/null @@ -1,57 +0,0 @@ -/** - * [53] Maximum Subarray - * - * Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. - * - * Example: - * - * - * Input: [-2,1,-3,4,-1,2,1,-5,4], - * Output: 6 - * Explanation: [4,-1,2,1] has the largest sum = 6. - * - * - * Follow up: - * - * If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/maximum-subarray/ -// discuss: https://leetcode.com/problems/maximum-subarray/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn max_sub_array(nums: Vec) -> i32 { - let mut j = 0_usize; - let mut max = i32::min_value(); - let mut curr = 0; - for j in 0..nums.len() { - curr += nums[j]; - max = i32::max(max, curr); - if curr <= 0 { - curr = 0; - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_53() { - assert_eq!( - Solution::max_sub_array(vec![-2, 1, -3, 4, -1, 2, 1, -5, 4]), - 6 - ); - assert_eq!(Solution::max_sub_array(vec![-8]), -8); - assert_eq!(Solution::max_sub_array(vec![-8, -2]), -2); - } -} diff --git a/src/solution/s0054_spiral_matrix.rs b/src/solution/s0054_spiral_matrix.rs deleted file mode 100644 index ac90b769..00000000 --- a/src/solution/s0054_spiral_matrix.rs +++ /dev/null @@ -1,101 +0,0 @@ -/** - * [54] Spiral Matrix - * - * Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. - * - * Example 1: - * - * - * Input: - * [ - * [ 1, 2, 3 ], - * [ 4, 5, 6 ], - * [ 7, 8, 9 ] - * ] - * Output: [1,2,3,6,9,8,7,4,5] - * - * - * Example 2: - * - * Input: - * [ - * [1, 2, 3, 4], - * [5, 6, 7, 8], - * [9,10,11,12] - * ] - * Output: [1,2,3,4,8,12,11,10,9,5,6,7] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/spiral-matrix/ -// discuss: https://leetcode.com/problems/spiral-matrix/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn spiral_order(matrix: Vec>) -> Vec { - let mut res = Vec::new(); - if matrix.len() < 1 { - return res; - } - let (height, width) = (matrix.len(), matrix[0].len()); - let (mut x_min, mut x_max, mut y_min, mut y_max) = (0, height, 0, width); - loop { - for y in y_min..y_max { - res.push(matrix[x_min][y]) - } - x_min += 1; - if x_min == x_max { - break; - } - for x in x_min..x_max { - res.push(matrix[x][y_max - 1]) - } - y_max -= 1; - if y_min == y_max { - break; - } - for y in (y_min..y_max).rev() { - res.push(matrix[x_max - 1][y]) - } - x_max -= 1; - if x_min == x_max { - break; - } - for x in (x_min..x_max).rev() { - res.push(matrix[x][y_min]) - } - y_min += 1; - if y_min == y_max { - break; - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_54() { - assert_eq!( - Solution::spiral_order(vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]]), - vec![1, 2, 3, 6, 9, 8, 7, 4, 5] - ); - assert_eq!(Solution::spiral_order(vec![vec![1, 2, 3]]), vec![1, 2, 3]); - assert_eq!( - Solution::spiral_order(vec![vec![1], vec![2], vec![3],]), - vec![1, 2, 3] - ); - assert_eq!(Solution::spiral_order(vec![vec![1],]), vec![1]); - assert_eq!( - Solution::spiral_order(vec![vec![1, 2], vec![4, 5],]), - vec![1, 2, 5, 4] - ); - } -} diff --git a/src/solution/s0055_jump_game.rs b/src/solution/s0055_jump_game.rs deleted file mode 100644 index 8dc35842..00000000 --- a/src/solution/s0055_jump_game.rs +++ /dev/null @@ -1,66 +0,0 @@ -/** - * [55] Jump Game - * - * Given an array of non-negative integers, you are initially positioned at the first index of the array. - * - * Each element in the array represents your maximum jump length at that position. - * - * Determine if you are able to reach the last index. - * - * Example 1: - * - * - * Input: [2,3,1,1,4] - * Output: true - * Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index. - * - * - * Example 2: - * - * - * Input: [3,2,1,0,4] - * Output: false - * Explanation: You will always arrive at index 3 no matter what. Its maximum - * jump length is 0, which makes it impossible to reach the last index. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/jump-game/ -// discuss: https://leetcode.com/problems/jump-game/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn can_jump(nums: Vec) -> bool { - let mut max_idx = 0_usize; - let mut start = 0_usize; - while start < nums.len() && start <= max_idx { - max_idx = usize::max(start + nums[start] as usize, max_idx); - start += 1; - } - return max_idx >= (nums.len() - 1); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_55() { - assert_eq!(Solution::can_jump(vec![2, 3, 1, 1, 4]), true); - assert_eq!(Solution::can_jump(vec![3, 2, 1, 0, 4]), false); - assert_eq!(Solution::can_jump(vec![2, 3, 1, 1, 0, 0, 0, 4]), false); - assert_eq!(Solution::can_jump(vec![8, 3, 1, 1, 0, 0, 0, 4]), true); - assert_eq!(Solution::can_jump(vec![0]), true); - assert_eq!(Solution::can_jump(vec![1, 1, 2, 2, 0, 1, 1]), true); - assert_eq!( - Solution::can_jump(vec![1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0]), - true - ); - } -} diff --git a/src/solution/s0056_merge_intervals.rs b/src/solution/s0056_merge_intervals.rs deleted file mode 100644 index 108c043b..00000000 --- a/src/solution/s0056_merge_intervals.rs +++ /dev/null @@ -1,85 +0,0 @@ -/** - * [56] Merge Intervals - * - * Given a collection of intervals, merge all overlapping intervals. - * - * Example 1: - * - * - * Input: [[1,3],[2,6],[8,10],[15,18]] - * Output: [[1,6],[8,10],[15,18]] - * Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6]. - * - * - * Example 2: - * - * - * Input: [[1,4],[4,5]] - * Output: [[1,5]] - * Explanation: Intervals [1,4] and [4,5] are considered overlapping. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/merge-intervals/ -// discuss: https://leetcode.com/problems/merge-intervals/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for an interval. -#[derive(Debug, PartialEq, Eq)] -pub struct Interval { - pub start: i32, - pub end: i32, -} - -impl Interval { - #[inline] - pub fn new(start: i32, end: i32) -> Self { - Interval { start, end } - } -} -impl Solution { - pub fn merge(intervals: Vec) -> Vec { - let mut intervals = intervals; - intervals.sort_unstable_by_key(|interval| interval.start); - let mut result: Vec = Vec::new(); - for interval in intervals.into_iter() { - match result.last_mut() { - Some(mut last_inter) => { - if last_inter.end >= interval.start { - last_inter.end = i32::max(last_inter.end, interval.end); - continue; - } - } - None => {} - } - result.push(interval); - } - result - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_56() { - assert_eq!( - Solution::merge(vec![ - Interval::new(1, 3), - Interval::new(2, 6), - Interval::new(8, 10), - Interval::new(15, 18) - ]), - vec![ - Interval::new(1, 6), - Interval::new(8, 10), - Interval::new(15, 18) - ] - ); - } -} diff --git a/src/solution/s0057_insert_interval.rs b/src/solution/s0057_insert_interval.rs deleted file mode 100644 index 0afe62b4..00000000 --- a/src/solution/s0057_insert_interval.rs +++ /dev/null @@ -1,126 +0,0 @@ -/** - * [57] Insert Interval - * - * Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). - * - * You may assume that the intervals were initially sorted according to their start times. - * - * Example 1: - * - * - * Input: intervals = [[1,3],Interval::new(6,9)], newInterval = [2,5] - * Output: [[1,5],[6,9]] - * - * - * Example 2: - * - * - * Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8] - * Output: [[1,2],[3,10],[12,16]] - * Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10]. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/insert-interval/ -// discuss: https://leetcode.com/problems/insert-interval/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for an interval. -#[derive(Debug, PartialEq, Eq)] -pub struct Interval { - pub start: i32, - pub end: i32, -} - -impl Interval { - #[inline] - pub fn new(start: i32, end: i32) -> Self { - Interval { start, end } - } -} -impl Solution { - pub fn insert(intervals: Vec, new_interval: Interval) -> Vec { - let mut result = Vec::new(); - let mut new_interval = new_interval; - let mut inserting = false; - let mut inserted = false; - for interval in intervals.into_iter() { - if new_interval.start <= interval.end && !inserted { - inserting = true; - } - if inserting { - if new_interval.end >= interval.start { - new_interval.start = i32::min(new_interval.start, interval.start); - new_interval.end = i32::max(new_interval.end, interval.end); - } else { - result.push(Interval::new(new_interval.start, new_interval.end)); - inserting = false; - inserted = true; - } - } - if !inserting { - result.push(interval); - } - } - if !inserted { - result.push(new_interval); - } - result - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_57() { - assert_eq!( - Solution::insert( - vec![Interval::new(1, 3), Interval::new(6, 9)], - Interval::new(2, 5) - ), - vec![Interval::new(1, 5), Interval::new(6, 9)] - ); - assert_eq!( - Solution::insert( - vec![ - Interval::new(1, 2), - Interval::new(3, 5), - Interval::new(6, 7), - Interval::new(8, 10), - Interval::new(12, 16) - ], - Interval::new(4, 8) - ), - vec![ - Interval::new(1, 2), - Interval::new(3, 10), - Interval::new(12, 16) - ] - ); - assert_eq!( - Solution::insert(vec![Interval::new(3, 4)], Interval::new(1, 2)), - vec![Interval::new(1, 2), Interval::new(3, 4)] - ); - assert_eq!( - Solution::insert(vec![Interval::new(1, 2)], Interval::new(3, 4)), - vec![Interval::new(1, 2), Interval::new(3, 4)] - ); - assert_eq!( - Solution::insert(vec![Interval::new(1, 2)], Interval::new(2, 3)), - vec![Interval::new(1, 3)] - ); - assert_eq!( - Solution::insert( - vec![Interval::new(1, 2), Interval::new(3, 4)], - Interval::new(0, 6) - ), - vec![Interval::new(0, 6)] - ); - } -} diff --git a/src/solution/s0058_length_of_last_word.rs b/src/solution/s0058_length_of_last_word.rs deleted file mode 100644 index 4889c739..00000000 --- a/src/solution/s0058_length_of_last_word.rs +++ /dev/null @@ -1,55 +0,0 @@ -/** - * [58] Length of Last Word - * - * Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. - * - * If the last word does not exist, return 0. - * - * Note: A word is defined as a character sequence consists of non-space characters only. - * - * Example: - * - * Input: "Hello World" - * Output: 5 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/length-of-last-word/ -// discuss: https://leetcode.com/problems/length-of-last-word/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn length_of_last_word(s: String) -> i32 { - let seq: Vec = s.chars().rev().collect(); - let mut result = 0; - let mut find = false; - for ch in seq { - if ch == ' ' && find { - break; - } - if ch != ' ' { - find = true; - result += 1; - } - } - result - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_58() { - assert_eq!(Solution::length_of_last_word("Hello World".to_owned()), 5); - assert_eq!(Solution::length_of_last_word(" ".to_owned()), 0); - assert_eq!(Solution::length_of_last_word("".to_owned()), 0); - assert_eq!(Solution::length_of_last_word(" rrrrr ".to_owned()), 5); - } -} diff --git a/src/solution/s0059_spiral_matrix_ii.rs b/src/solution/s0059_spiral_matrix_ii.rs deleted file mode 100644 index d54f57b4..00000000 --- a/src/solution/s0059_spiral_matrix_ii.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [59] Spiral Matrix II - * - * Given a positive integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order. - * - * Example: - * - * - * Input: 3 - * Output: - * [ - * [ 1, 2, 3 ], - * [ 8, 9, 4 ], - * [ 7, 6, 5 ] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/spiral-matrix-ii/ -// discuss: https://leetcode.com/problems/spiral-matrix-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn generate_matrix(n: i32) -> Vec> { - let mut res = vec![vec![0; n as usize]; n as usize]; - if n < 1 { - return res; - } - let (mut x_min, mut x_max, mut y_min, mut y_max) = (0, n as usize, 0, n as usize); - let mut i = 1; - loop { - for y in y_min..y_max { - res[x_min][y] = i; - i += 1; - } - x_min += 1; - if x_min == x_max { - break; - } - for x in x_min..x_max { - res[x][y_max - 1] = i; - i += 1; - } - y_max -= 1; - if y_min == y_max { - break; - } - for y in (y_min..y_max).rev() { - res[x_max - 1][y] = i; - i += 1; - } - x_max -= 1; - if x_min == x_max { - break; - } - for x in (x_min..x_max).rev() { - res[x][y_min] = i; - i += 1; - } - y_min += 1; - if y_min == y_max { - break; - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_59() { - assert_eq!(Solution::generate_matrix(1), vec![vec![1]]); - assert_eq!(Solution::generate_matrix(2), vec![vec![1, 2], vec![4, 3]]); - assert_eq!( - Solution::generate_matrix(3), - vec![vec![1, 2, 3], vec![8, 9, 4], vec![7, 6, 5],] - ); - } -} diff --git a/src/solution/s0060_permutation_sequence.rs b/src/solution/s0060_permutation_sequence.rs deleted file mode 100644 index 0136a0b4..00000000 --- a/src/solution/s0060_permutation_sequence.rs +++ /dev/null @@ -1,76 +0,0 @@ -/** - * [60] Permutation Sequence - * - * The set [1,2,3,...,n] contains a total of n! unique permutations. - * - * By listing and labeling all of the permutations in order, we get the following sequence for n = 3: - * - *
    - * "123" - * "132" - * "213" - * "231" - * "312" - * "321" - *
- * - * Given n and k, return the k^th permutation sequence. - * - * Note: - * - * - * Given n will be between 1 and 9 inclusive. - * Given k will be between 1 and n! inclusive. - * - * - * Example 1: - * - * - * Input: n = 3, k = 3 - * Output: "213" - * - * - * Example 2: - * - * - * Input: n = 4, k = 9 - * Output: "2314" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/permutation-sequence/ -// discuss: https://leetcode.com/problems/permutation-sequence/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// target: split k = i! + j! + ... -use std::char::from_digit; -impl Solution { - pub fn get_permutation(n: i32, k: i32) -> String { - let factorials = [0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]; - let mut k = k; - let mut i = n; - let mut res = String::new(); - while i > 0 { - if k > factorials[i as usize] { - let round = k / factorials[i as usize]; - if round >= n {} - } else { - i -= 1; - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_60() {} -} diff --git a/src/solution/s0061_rotate_list.rs b/src/solution/s0061_rotate_list.rs deleted file mode 100644 index bbfb059f..00000000 --- a/src/solution/s0061_rotate_list.rs +++ /dev/null @@ -1,50 +0,0 @@ -/** - * [61] Rotate List - * - * Given a linked list, rotate the list to the right by k places, where k is non-negative. - * - * Example 1: - * - * - * Input: 1->2->3->4->5->NULL, k = 2 - * Output: 4->5->1->2->3->NULL - * Explanation: - * rotate 1 steps to the right: 5->1->2->3->4->NULL - * rotate 2 steps to the right: 4->5->1->2->3->NULL - * - * - * Example 2: - * - * - * Input: 0->1->2->NULL, k = 4 - * Output: 2->0->1->NULL - * Explanation: - * rotate 1 steps to the right: 2->0->1->NULL - * rotate 2 steps to the right: 1->2->0->NULL - * rotate 3 steps to the right: 0->1->2->NULL - * rotate 4 steps to the right: 2->0->1->NULL - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/rotate-list/ -// discuss: https://leetcode.com/problems/rotate-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn rotate_right(head: Option>, k: i32) -> Option> { - None - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_61() {} -} diff --git a/src/solution/s0062_unique_paths.rs b/src/solution/s0062_unique_paths.rs deleted file mode 100644 index 3a158326..00000000 --- a/src/solution/s0062_unique_paths.rs +++ /dev/null @@ -1,80 +0,0 @@ -/** - * [62] Unique Paths - * - * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). - * - * The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). - * - * How many possible unique paths are there? - * - *
- * Above is a 7 x 3 grid. How many possible unique paths are there? - * - * Note: m and n will be at most 100. - * - * Example 1: - * - * - * Input: m = 3, n = 2 - * Output: 3 - * Explanation: - * From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: - * 1. Right -> Right -> Down - * 2. Right -> Down -> Right - * 3. Down -> Right -> Right - * - * - * Example 2: - * - * - * Input: m = 7, n = 3 - * Output: 28 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/unique-paths/ -// discuss: https://leetcode.com/problems/unique-paths/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// its high school math: C(r,n) = n! / r!(n-r)! ...are you fxxking kidding me? -// ...high school math will attempt to i32 overflow, we have to do it clever -impl Solution { - pub fn unique_paths(m: i32, n: i32) -> i32 { - let (m, n) = ((m - 1) as u64, (n - 1) as u64); - let sum = m + n; - (Solution::partial_factorial(u64::max(m, n), sum) - / Solution::partial_factorial(0, u64::min(m, n))) as i32 - } - - #[inline(always)] - pub fn partial_factorial(start: u64, mut end: u64) -> u64 { - if start > end { - unreachable!() - } - let mut res = 1; - while end > start { - println!("{}", end); - res *= end; - end -= 1; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_62() { - assert_eq!(Solution::unique_paths(7, 3), 28); - assert_eq!(Solution::unique_paths(3, 7), 28); - assert_eq!(Solution::unique_paths(1, 1), 1); - assert_eq!(Solution::unique_paths(2, 2), 2); - assert_eq!(Solution::unique_paths(36, 7), 4496388); - } -} diff --git a/src/solution/s0063_unique_paths_ii.rs b/src/solution/s0063_unique_paths_ii.rs deleted file mode 100644 index 04d84911..00000000 --- a/src/solution/s0063_unique_paths_ii.rs +++ /dev/null @@ -1,111 +0,0 @@ -/** - * [63] Unique Paths II - * - * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). - * - * The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). - * - * Now consider if some obstacles are added to the grids. How many unique paths would there be? - * - * - * - * An obstacle and empty space is marked as 1 and 0 respectively in the grid. - * - * Note: m and n will be at most 100. - * - * Example 1: - * - * - * Input: - * [ - * [0,0,0], - * [0,1,0], - * [0,0,0] - * ] - * Output: 2 - * Explanation: - * There is one obstacle in the middle of the 3x3 grid above. - * There are two ways to reach the bottom-right corner: - * 1. Right -> Right -> Down -> Down - * 2. Down -> Down -> Right -> Right - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/unique-paths-ii/ -// discuss: https://leetcode.com/problems/unique-paths-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Bottom-Up DP -impl Solution { - pub fn unique_paths_with_obstacles(obstacle_grid: Vec>) -> i32 { - let (height, width) = (obstacle_grid.len(), obstacle_grid[0].len()); - let mut paths: Vec> = vec![vec![0; width]; height]; - if obstacle_grid[0][0] != 1 { - paths[0][0] = 1; - } - let mut step = 1; - while step <= height + width - 2 { - for x in 0..(step + 1) { - let y = step - x; - if x >= height || y >= width || obstacle_grid[x][y] == 1 { - continue; - } - if y >= 1 { - paths[x][y] = paths[x][y] + paths[x][y - 1]; - } - if x >= 1 { - paths[x][y] = paths[x][y] + paths[x - 1][y]; - } - } - step += 1; - } - paths[height - 1][width - 1] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_63() { - assert_eq!(Solution::unique_paths_with_obstacles(vec![vec![0]]), 1); - assert_eq!( - Solution::unique_paths_with_obstacles(vec![vec![0, 0], vec![0, 0],]), - 2 - ); - assert_eq!( - Solution::unique_paths_with_obstacles(vec![vec![0, 1], vec![1, 0],]), - 0 - ); - assert_eq!( - Solution::unique_paths_with_obstacles(vec![ - vec![0, 0, 0], - vec![0, 1, 0], - vec![0, 0, 0], - ]), - 2 - ); - assert_eq!( - Solution::unique_paths_with_obstacles(vec![ - vec![0, 0, 0, 0], - vec![0, 0, 0, 0], - vec![0, 0, 0, 0], - ]), - 10 - ); - assert_eq!( - Solution::unique_paths_with_obstacles(vec![ - vec![0, 0, 0, 0], - vec![0, 0, 0, 1], - vec![0, 0, 1, 0], - ]), - 0 - ); - } -} diff --git a/src/solution/s0064_minimum_path_sum.rs b/src/solution/s0064_minimum_path_sum.rs deleted file mode 100644 index d6340d50..00000000 --- a/src/solution/s0064_minimum_path_sum.rs +++ /dev/null @@ -1,69 +0,0 @@ -/** - * [64] Minimum Path Sum - * - * Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. - * - * Note: You can only move either down or right at any point in time. - * - * Example: - * - * - * Input: - * [ - * [1,3,1], - * [1,5,1], - * [4,2,1] - * ] - * Output: 7 - * Explanation: Because the path 1→3→1→1→1 minimizes the sum. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/minimum-path-sum/ -// discuss: https://leetcode.com/problems/minimum-path-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn min_path_sum(grid: Vec>) -> i32 { - let (height, width) = (grid.len(), grid[0].len()); - let mut grid = grid; - let mut step = 1; - while step <= height + width - 2 { - for x in 0..(step + 1) { - let y = step - x; - if x >= height || y >= width { - continue; - } - if x < 1 { - grid[x][y] += grid[x][y - 1]; - } else if y < 1 { - grid[x][y] += grid[x - 1][y]; - } else { - grid[x][y] += i32::min(grid[x][y - 1], grid[x - 1][y]); - } - } - step += 1; - } - grid[height - 1][width - 1] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_64() { - assert_eq!(Solution::min_path_sum(vec![vec![2]]), 2); - assert_eq!( - Solution::min_path_sum(vec![vec![1, 3, 1], vec![1, 5, 1], vec![4, 2, 1],]), - 7 - ); - assert_eq!(Solution::min_path_sum(vec![vec![1, 3, 1],]), 5); - } -} diff --git a/src/solution/s0065_valid_number.rs b/src/solution/s0065_valid_number.rs deleted file mode 100644 index 11c5753d..00000000 --- a/src/solution/s0065_valid_number.rs +++ /dev/null @@ -1,60 +0,0 @@ -/** - * [65] Valid Number - * - * Validate if a given string can be interpreted as a decimal number. - * - * Some examples:
- * "0" => true
- * " 0.1 " => true
- * "abc" => false
- * "1 a" => false
- * "2e10" => true
- * " -90e3 " => true
- * " 1e" => false
- * "e3" => false
- * " 6e-1" => true
- * " 99e2.5 " => false
- * "53.5e93" => true
- * " --6 " => false
- * "-+3" => false
- * "95a54e53" => false - * - * Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number: - * - * - * Numbers 0-9 - * Exponent - "e" - * Positive/negative sign - "+"/"-" - * Decimal point - "." - * - * - * Of course, the context of these characters also matters in the input. - * - * Update (2015-02-10):
- * The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/valid-number/ -// discuss: https://leetcode.com/problems/valid-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// hope that regex was included in std library... -// TODO: NFA -impl Solution { - pub fn is_number(s: String) -> bool { - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_65() {} -} diff --git a/src/solution/s0066_plus_one.rs b/src/solution/s0066_plus_one.rs deleted file mode 100644 index fb1ea045..00000000 --- a/src/solution/s0066_plus_one.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [66] Plus One - * - * Given a non-empty array of digits representing a non-negative integer, plus one to the integer. - * - * The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit. - * - * You may assume the integer does not contain any leading zero, except the number 0 itself. - * - * Example 1: - * - * - * Input: [1,2,3] - * Output: [1,2,4] - * Explanation: The array represents the integer 123. - * - * - * Example 2: - * - * - * Input: [4,3,2,1] - * Output: [4,3,2,2] - * Explanation: The array represents the integer 4321. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/plus-one/ -// discuss: https://leetcode.com/problems/plus-one/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn plus_one(digits: Vec) -> Vec { - let mut digits = digits; - let mut carry = 0; - for i in (0..digits.len()).rev() { - digits[i] = if digits[i] == 9 { - carry = 1; - 0 - } else { - carry = 0; - digits[i] + 1 - }; - if carry == 0 { - break; - } - } - if carry > 0 { - digits.insert(0, 1); - } - digits - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_66() { - assert_eq!(Solution::plus_one(vec![0]), vec![1]); - assert_eq!(Solution::plus_one(vec![9, 9, 9, 9]), vec![1, 0, 0, 0, 0]); - assert_eq!( - Solution::plus_one(vec![1, 0, 9, 9, 9, 9]), - vec![1, 1, 0, 0, 0, 0] - ); - } -} diff --git a/src/solution/s0067_add_binary.rs b/src/solution/s0067_add_binary.rs deleted file mode 100644 index 8c9951cc..00000000 --- a/src/solution/s0067_add_binary.rs +++ /dev/null @@ -1,79 +0,0 @@ -/** - * [67] Add Binary - * - * Given two binary strings, return their sum (also a binary string). - * - * The input strings are both non-empty and contains only characters 1 or 0. - * - * Example 1: - * - * - * Input: a = "11", b = "1" - * Output: "100" - * - * Example 2: - * - * - * Input: a = "1010", b = "1011" - * Output: "10101" - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/add-binary/ -// discuss: https://leetcode.com/problems/add-binary/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::char::from_digit; -impl Solution { - pub fn add_binary(a: String, b: String) -> String { - let mut buf = Vec::with_capacity(usize::max(a.len(), b.len()) + 1); - let mut a: Vec = a.chars().collect(); - let mut b: Vec = b.chars().collect(); - let mut carry = 0; - while !(a.is_empty() && b.is_empty()) { - let mut sum = a.pop().map_or(0, |ch| ch.to_digit(10).unwrap()) - + b.pop().map_or(0, |ch| ch.to_digit(10).unwrap()) - + carry; - if sum > 1 { - sum -= 2; - carry = 1; - } else { - carry = 0; - } - buf.push(from_digit(sum, 10).unwrap()) - } - if carry > 0 { - buf.push('1') - } - buf.into_iter().rev().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_67() { - assert_eq!( - Solution::add_binary("0".to_owned(), "0".to_owned()), - "0".to_owned() - ); - assert_eq!( - Solution::add_binary("1010".to_owned(), "1011".to_owned()), - "10101".to_owned() - ); - assert_eq!( - Solution::add_binary("11".to_owned(), "1".to_owned()), - "100".to_owned() - ); - assert_eq!( - Solution::add_binary("1111".to_owned(), "1111".to_owned()), - "11110".to_owned() - ); - } -} diff --git a/src/solution/s0068_text_justification.rs b/src/solution/s0068_text_justification.rs deleted file mode 100644 index 827fa4f8..00000000 --- a/src/solution/s0068_text_justification.rs +++ /dev/null @@ -1,200 +0,0 @@ -/** - * [68] Text Justification - * - * Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified. - * - * You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters. - * - * Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right. - * - * For the last line of text, it should be left justified and no extra space is inserted between words. - * - * Note: - * - * - * A word is defined as a character sequence consisting of non-space characters only. - * Each word's length is guaranteed to be greater than 0 and not exceed maxWidth. - * The input array words contains at least one word. - * - * - * Example 1: - * - * - * Input: - * words = ["This", "is", "an", "example", "of", "text", "justification."] - * maxWidth = 16 - * Output: - * [ - * "This is an", - * "example of text", - * "justification. " - * ] - * - * - * Example 2: - * - * - * Input: - * words = ["What","must","be","acknowledgment","shall","be"] - * maxWidth = 16 - * Output: - * [ - * "What must be", - * "acknowledgment ", - * "shall be " - * ] - * Explanation: Note that the last line is "shall be " instead of "shall be", - * because the last line must be left-justified instead of fully-justified. - * Note that the second line is also left-justified becase it contains only one word. - * - * - * Example 3: - * - * - * Input: - * words = ["Science","is","what","we","understand","well","enough","to","explain", - * "to","a","computer.","Art","is","everything","else","we","do"] - * maxWidth = 20 - * Output: - * [ - * "Science is what we", - * "understand well", - * "enough to explain to", - * "a computer. Art is", - * "everything else we", - * "do " - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/text-justification/ -// discuss: https://leetcode.com/problems/text-justification/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn full_justify(words: Vec, max_width: i32) -> Vec { - let mut res = Vec::new(); - let max_width = max_width as usize; - let mut i = 0; - let mut row_len = 0; - let mut buf = Vec::new(); - while i < words.len() { - if words[i].len() > max_width { - unreachable!() - } - let old_len = row_len; - row_len += words[i].len() + if row_len > 0 { 1 } else { 0 }; - if row_len > max_width { - res.push(Solution::compact(buf, max_width, old_len)); - buf = Vec::new(); - row_len = 0; - } else { - buf.push(words[i].clone()); - i += 1; - } - } - res.push(Solution::compact_last(buf, max_width)); - res - } - - fn compact(words: Vec, max_width: usize, row_len: usize) -> String { - let spaces = max_width - (row_len - words.len() + 1); - let avg_spaces = spaces / usize::max(1, words.len() - 1); - let mut extra_spaces = spaces - avg_spaces * usize::max(1, words.len() - 1);; - let mut res = String::new(); - for (i, word) in words.iter().enumerate() { - res.push_str(word); - if words.len() < 2 || (i < words.len() - 1) { - res.push_str(&" ".repeat(avg_spaces)); - if extra_spaces > 0 { - res.push(' '); - extra_spaces -= 1; - } - } - } - res - } - - fn compact_last(words: Vec, max_width: usize) -> String { - let mut res = String::new(); - for (i, word) in words.iter().enumerate() { - res.push_str(word); - if i < words.len() - 1 { - res.push(' '); - } - } - res.push_str(&" ".repeat(max_width - res.len())); - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_68() { - assert_eq!( - Solution::full_justify( - vec_string![ - "This", - "is", - "an", - "example", - "of", - "text", - "justification." - ], - 16 - ), - vec_string!["This is an", "example of text", "justification. "] - ); - - assert_eq!( - Solution::full_justify( - vec_string!["What", "must", "be", "acknowledgment", "shall", "be"], - 16 - ), - vec_string!["What must be", "acknowledgment ", "shall be "] - ); - - assert_eq!( - Solution::full_justify( - vec_string![ - "Science", - "is", - "what", - "we", - "understand", - "well", - "enough", - "to", - "explain", - "to", - "a", - "computer.", - "Art", - "is", - "everything", - "else", - "we", - "do" - ], - 20 - ), - vec_string![ - "Science is what we", - "understand well", - "enough to explain to", - "a computer. Art is", - "everything else we", - "do ", - ] - ); - } -} diff --git a/src/solution/s0069_sqrtx.rs b/src/solution/s0069_sqrtx.rs deleted file mode 100644 index 256f88be..00000000 --- a/src/solution/s0069_sqrtx.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [69] Sqrt(x) - * - * Implement int sqrt(int x). - * - * Compute and return the square root of x, where x is guaranteed to be a non-negative integer. - * - * Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned. - * - * Example 1: - * - * - * Input: 4 - * Output: 2 - * - * - * Example 2: - * - * - * Input: 8 - * Output: 2 - * Explanation: The square root of 8 is 2.82842..., and since - * the decimal part is truncated, 2 is returned. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/sqrtx/ -// discuss: https://leetcode.com/problems/sqrtx/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Newton-Raphson for: root^2 - n = 0 -// Tangent equation: y = 2 * root * x - (root^2 + n) -// Zero point: (root^2 + n) / (2 * root) -impl Solution { - pub fn my_sqrt(x: i32) -> i32 { - let mut size = x; - let mut base = 1; - while size > 1 { - let half = size / 2; - let mid = base + half; - if mid <= x / mid { - base = mid; - } - size -= half; - } - base - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_69() { - assert_eq!(Solution::my_sqrt(8), 2); - assert_eq!(Solution::my_sqrt(16), 4); - assert_eq!(Solution::my_sqrt(17), 4); - assert_eq!(Solution::my_sqrt(81), 9); - assert_eq!(Solution::my_sqrt(82), 9); - assert_eq!(Solution::my_sqrt(100480577), 10024); - assert_eq!(Solution::my_sqrt(100480575), 10023); - assert_eq!(Solution::my_sqrt(100480575), 10023); - assert_eq!(Solution::my_sqrt(80), 8); - assert_eq!(Solution::my_sqrt(2), 1); - } -} diff --git a/src/solution/s0070_climbing_stairs.rs b/src/solution/s0070_climbing_stairs.rs deleted file mode 100644 index 2ef4153b..00000000 --- a/src/solution/s0070_climbing_stairs.rs +++ /dev/null @@ -1,71 +0,0 @@ -/** - * [70] Climbing Stairs - * - * You are climbing a stair case. It takes n steps to reach to the top. - * - * Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? - * - * Note: Given n will be a positive integer. - * - * Example 1: - * - * - * Input: 2 - * Output: 2 - * Explanation: There are two ways to climb to the top. - * 1. 1 step + 1 step - * 2. 2 steps - * - * - * Example 2: - * - * - * Input: 3 - * Output: 3 - * Explanation: There are three ways to climb to the top. - * 1. 1 step + 1 step + 1 step - * 2. 1 step + 2 steps - * 3. 2 steps + 1 step - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/climbing-stairs/ -// discuss: https://leetcode.com/problems/climbing-stairs/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Bottom-up DP -impl Solution { - pub fn climb_stairs(n: i32) -> i32 { - let n = n as usize; - if n == 1 { - return 1; - } - if n == 2 { - return 2; - } - let (mut prev, mut curr) = (1, 2); - for i in 2..n { - let next = prev + curr; - prev = curr; - curr = next; - } - curr - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_70() { - assert_eq!(Solution::climb_stairs(3), 3); - assert_eq!(Solution::climb_stairs(4), 5); - assert_eq!(Solution::climb_stairs(5), 8); - } -} diff --git a/src/solution/s0071_simplify_path.rs b/src/solution/s0071_simplify_path.rs deleted file mode 100644 index 4b61f7be..00000000 --- a/src/solution/s0071_simplify_path.rs +++ /dev/null @@ -1,112 +0,0 @@ -/** - * [71] Simplify Path - * - * Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path. - * - * In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix - * - * Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path. - * - * - * - * Example 1: - * - * - * Input: "/home/" - * Output: "/home" - * Explanation: Note that there is no trailing slash after the last directory name. - * - * - * Example 2: - * - * - * Input: "/../" - * Output: "/" - * Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go. - * - * - * Example 3: - * - * - * Input: "/home//foo/" - * Output: "/home/foo" - * Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one. - * - * - * Example 4: - * - * - * Input: "/a/./b/../../c/" - * Output: "/c" - * - * - * Example 5: - * - * - * Input: "/a/../../b/../c//.//" - * Output: "/c" - * - * - * Example 6: - * - * - * Input: "/a//b////c/d//././/.." - * Output: "/a/b/c" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/simplify-path/ -// discuss: https://leetcode.com/problems/simplify-path/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn simplify_path(path: String) -> String { - let mut stack = Vec::new(); - for s in path.split('/') { - match s { - "." => {} - "/" => {} - "" => {} - ".." => { - stack.pop(); - } - _ => stack.push(s), - } - } - let mut res = String::new(); - for s in stack { - res.push('/'); - res.push_str(s); - } - if res.len() > 0 { - res - } else { - "/".to_owned() - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_71() { - assert_eq!(Solution::simplify_path("/home/".to_owned()), "/home"); - assert_eq!(Solution::simplify_path("/../".to_owned()), "/"); - assert_eq!(Solution::simplify_path("/a/./b/../../c/".to_owned()), "/c"); - assert_eq!( - Solution::simplify_path("/a/../../b/../c//.//".to_owned()), - "/c" - ); - assert_eq!( - Solution::simplify_path("/a//b////c/d//././/..".to_owned()), - "/a/b/c" - ); - } -} diff --git a/src/solution/s0072_edit_distance.rs b/src/solution/s0072_edit_distance.rs deleted file mode 100644 index d50d9fa3..00000000 --- a/src/solution/s0072_edit_distance.rs +++ /dev/null @@ -1,60 +0,0 @@ -/** - * [72] Edit Distance - * - * Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. - * - * You have the following 3 operations permitted on a word: - * - *
    - * Insert a character - * Delete a character - * Replace a character - *
- * - * Example 1: - * - * - * Input: word1 = "horse", word2 = "ros" - * Output: 3 - * Explanation: - * horse -> rorse (replace 'h' with 'r') - * rorse -> rose (remove 'r') - * rose -> ros (remove 'e') - * - * - * Example 2: - * - * - * Input: word1 = "intention", word2 = "execution" - * Output: 5 - * Explanation: - * intention -> inention (remove 't') - * inention -> enention (replace 'i' with 'e') - * enention -> exention (replace 'n' with 'x') - * exention -> exection (replace 'n' with 'c') - * exection -> execution (insert 'u') - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/edit-distance/ -// discuss: https://leetcode.com/problems/edit-distance/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn min_distance(word1: String, word2: String) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_72() {} -} diff --git a/src/solution/s0073_set_matrix_zeroes.rs b/src/solution/s0073_set_matrix_zeroes.rs deleted file mode 100644 index 8d8cdab8..00000000 --- a/src/solution/s0073_set_matrix_zeroes.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [73] Set Matrix Zeroes - * - * Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place. - * - * Example 1: - * - * - * Input: - * [ - * [1,1,1], - * [1,0,1], - * [1,1,1] - * ] - * Output: - * [ - * [1,0,1], - * [0,0,0], - * [1,0,1] - * ] - * - * - * Example 2: - * - * - * Input: - * [ - * [0,1,2,0], - * [3,4,5,2], - * [1,3,1,5] - * ] - * Output: - * [ - * [0,0,0,0], - * [0,4,5,0], - * [0,3,1,0] - * ] - * - * - * Follow up: - * - * - * A straight forward solution using O(mn) space is probably a bad idea. - * A simple improvement uses O(m + n) space, but still not the best solution. - * Could you devise a constant space solution? - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/set-matrix-zeroes/ -// discuss: https://leetcode.com/problems/set-matrix-zeroes/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn set_zeroes(matrix: &mut Vec>) {} -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_73() {} -} diff --git a/src/solution/s0074_search_a_2d_matrix.rs b/src/solution/s0074_search_a_2d_matrix.rs deleted file mode 100644 index 90975918..00000000 --- a/src/solution/s0074_search_a_2d_matrix.rs +++ /dev/null @@ -1,92 +0,0 @@ -/** - * [74] Search a 2D Matrix - * - * Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: - * - * - * Integers in each row are sorted from left to right. - * The first integer of each row is greater than the last integer of the previous row. - * - * - * Example 1: - * - * - * Input: - * matrix = [ - * [1, 3, 5, 7], - * [10, 11, 16, 20], - * [23, 30, 34, 50] - * ] - * target = 3 - * Output: true - * - * - * Example 2: - * - * - * Input: - * matrix = [ - * [1, 3, 5, 7], - * [10, 11, 16, 20], - * [23, 30, 34, 50] - * ] - * target = 13 - * Output: false - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/search-a-2d-matrix/ -// discuss: https://leetcode.com/problems/search-a-2d-matrix/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn search_matrix(matrix: Vec>, target: i32) -> bool { - if matrix.is_empty() { - return false; - } - let (height, width) = (matrix.len(), matrix[0].len()); - if height < 1 || width < 1 { - return false; - } - let mut size = height * width; - let mut base = 0_usize; - while size > 1 { - let half = size / 2; - let mid = base + half; - if target == matrix[mid / width][mid % width] { - return true; - } else if (target > matrix[mid / width][mid % width]) { - base = mid; - } - size -= half; - } - target == matrix[base / width][base % width] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_74() { - assert_eq!( - Solution::search_matrix( - vec![vec![1, 3, 5, 7], vec![10, 11, 16, 20], vec![23, 30, 34, 50]], - 3 - ), - true - ); - assert_eq!( - Solution::search_matrix( - vec![vec![1, 3, 5, 7], vec![10, 11, 16, 20], vec![23, 30, 34, 50]], - 13 - ), - false - ); - } -} diff --git a/src/solution/s0075_sort_colors.rs b/src/solution/s0075_sort_colors.rs deleted file mode 100644 index 6b72ac88..00000000 --- a/src/solution/s0075_sort_colors.rs +++ /dev/null @@ -1,84 +0,0 @@ -/** - * [75] Sort Colors - * - * Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue. - * - * Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively. - * - * Note: You are not suppose to use the library's sort function for this problem. - * - * Example: - * - * - * Input: [2,0,2,1,1,0] - * Output: [0,0,1,1,2,2] - * - * Follow up: - * - * - * A rather straight forward solution is a two-pass algorithm using counting sort.
- * First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's. - * Could you come up with a one-pass algorithm using only constant space? - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/sort-colors/ -// discuss: https://leetcode.com/problems/sort-colors/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// three-way partition -impl Solution { - pub fn sort_colors(nums: &mut Vec) { - if nums.is_empty() { - return; - } - let (mut lower_idx, mut upper_idx) = (0_usize, nums.len() - 1); - let mut i = 0_usize; - while i <= upper_idx { - if nums[i] < 1 { - // lower_idx <= i, we've scanned it so we know nums[lower_idx] <= 1, i++ - nums.swap(lower_idx, i); - i += 1; - lower_idx += 1; - } else if nums[i] > 1 { - nums.swap(upper_idx, i); - if upper_idx < 1 { - break; - } - upper_idx -= 1; - } else { - i += 1; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_75() { - let mut vec = vec![ - 1, 2, 0, 1, 2, 2, 2, 0, 0, 0, 2, 1, 1, 2, 0, 1, 2, 2, 1, 1, 0, - ]; - Solution::sort_colors(&mut vec); - assert_eq!( - vec, - vec![0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2] - ); - - let mut vec = vec![]; - Solution::sort_colors(&mut vec); - assert_eq!(vec, vec![]); - - let mut vec = vec![2, 2, 2]; - Solution::sort_colors(&mut vec); - assert_eq!(vec, vec![2, 2, 2]); - } -} diff --git a/src/solution/s0076_minimum_window_substring.rs b/src/solution/s0076_minimum_window_substring.rs deleted file mode 100644 index 9d190533..00000000 --- a/src/solution/s0076_minimum_window_substring.rs +++ /dev/null @@ -1,56 +0,0 @@ -/** - * [76] Minimum Window Substring - * - * Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). - * - * Example: - * - * - * Input: S = "ADOBECODEBANC", T = "ABC" - * Output: "BANC" - * - * - * Note: - * - * - * If there is no such window in S that covers all characters in T, return the empty string "". - * If there is such window, you are guaranteed that there will always be only one unique minimum window in S. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/minimum-window-substring/ -// discuss: https://leetcode.com/problems/minimum-window-substring/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -use std::collections::HashMap; -impl Solution { - pub fn min_window(s: String, t: String) -> String { - if t.is_empty() || t.len() > s.len() { - return "".to_owned(); - } - let (mut start, mut end) = (0_usize, 0_usize); - let mut result = (0_usize, 0_usize); - loop {} - s[result.0..result.1].to_owned() - } - - fn count_char(s: String) -> HashMap { - let mut res = HashMap::new(); - for ch in s.chars().into_iter() { - *res.entry(ch).or_insert(0) += 1; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_76() {} -} diff --git a/src/solution/s0077_combinations.rs b/src/solution/s0077_combinations.rs deleted file mode 100644 index 6bb10d6b..00000000 --- a/src/solution/s0077_combinations.rs +++ /dev/null @@ -1,77 +0,0 @@ -/** - * [77] Combinations - * - * Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. - * - * Example: - * - * - * Input: n = 4, k = 2 - * Output: - * [ - * [2,4], - * [3,4], - * [2,3], - * [1,2], - * [1,3], - * [1,4], - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/combinations/ -// discuss: https://leetcode.com/problems/combinations/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn combine(n: i32, k: i32) -> Vec> { - let mut res: Vec> = Vec::new(); - Solution::backtrack(1, n, k, vec![], &mut res); - res - } - - fn backtrack(start: i32, end: i32, k: i32, curr: Vec, result: &mut Vec>) { - if k < 1 { - result.push(curr); - return; - } - if end - start + 1 < k { - // elements is not enough, return quickly - return; - } - for i in start..end + 1 { - let mut vec = curr.clone(); - vec.push(i); - Solution::backtrack(i + 1, end, k - 1, vec, result); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_77() { - assert_eq!( - Solution::combine(4, 2), - vec![ - vec![1, 2], - vec![1, 3], - vec![1, 4], - vec![2, 3], - vec![2, 4], - vec![3, 4] - ] - ); - assert_eq!(Solution::combine(1, 1), vec![vec![1]]); - let empty: Vec> = vec![]; - assert_eq!(Solution::combine(0, 1), empty); - assert_eq!(Solution::combine(2, 1), vec![vec![1], vec![2]]); - } -} diff --git a/src/solution/s0078_subsets.rs b/src/solution/s0078_subsets.rs deleted file mode 100644 index 2db34925..00000000 --- a/src/solution/s0078_subsets.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [78] Subsets - * - * Given a set of distinct integers, nums, return all possible subsets (the power set). - * - * Note: The solution set must not contain duplicate subsets. - * - * Example: - * - * - * Input: nums = [1,2,3] - * Output: - * [ - * [3], - * [1], - * [2], - * [1,2,3], - * [1,3], - * [2,3], - * [1,2], - * [] - * ] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/subsets/ -// discuss: https://leetcode.com/problems/subsets/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn subsets(nums: Vec) -> Vec> { - let mut res = Vec::new(); - Solution::backtrack(0, vec![], &nums, &mut res); - res - } - - fn backtrack(start: usize, mut curr: Vec, nums: &Vec, result: &mut Vec>) { - if start >= nums.len() { - result.push(curr); - return; - } - // current element dropped - Solution::backtrack(start + 1, curr.clone(), nums, result); - // current element picked - curr.push(nums[start]); - Solution::backtrack(start + 1, curr, nums, result); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_78() { - assert_eq!(Solution::subsets(vec![]), vec![vec![]]); - assert_eq!(Solution::subsets(vec![1]), vec![vec![], vec![1]]); - assert_eq!( - Solution::subsets(vec![1, 2]), - vec![vec![], vec![2], vec![1], vec![1, 2]] - ); - } -} diff --git a/src/solution/s0079_word_search.rs b/src/solution/s0079_word_search.rs deleted file mode 100644 index 07c4a705..00000000 --- a/src/solution/s0079_word_search.rs +++ /dev/null @@ -1,133 +0,0 @@ -/** - * [79] Word Search - * - * Given a 2D board and a word, find if the word exists in the grid. - * - * The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. - * - * Example: - * - * - * board = - * [ - * ['A','B','C','E'], - * ['S','F','C','S'], - * ['A','D','E','E'] - * ] - * - * Given word = "ABCCED", return true. - * Given word = "SEE", return true. - * Given word = "ABCB", return false. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-search/ -// discuss: https://leetcode.com/problems/word-search/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: use HashSet to record visited pos -impl Solution { - pub fn exist(board: Vec>, word: String) -> bool { - if board.is_empty() || word.len() < 1 { - return false; - } - let (height, width) = (board.len(), board[0].len()); - if height < 1 || width < 1 { - return false; - } - let seq: Vec = word.chars().collect(); - - for i in 0..height * width { - if Solution::dfs( - i / width, - i % width, - &seq[..], - &board, - vec![], - height, - width, - ) { - return true; - } - } - false - } - - fn dfs( - x: usize, - y: usize, - seq: &[char], - board: &Vec>, - mut visited: Vec<(usize, usize)>, - height: usize, - width: usize, - ) -> bool { - if seq[0] != board[x][y] { - return false; - } - if seq.len() < 2 { - return true; - } - visited.push((x, y)); - return (x > 0 - && !visited.contains(&(x - 1, y)) - && Solution::dfs(x - 1, y, &seq[1..], board, visited.clone(), height, width)) - || (x + 1 < height - && !visited.contains(&(x + 1, y)) - && Solution::dfs(x + 1, y, &seq[1..], board, visited.clone(), height, width)) - || (y > 0 - && !visited.contains(&(x, y - 1)) - && Solution::dfs(x, y - 1, &seq[1..], board, visited.clone(), height, width)) - || (y + 1 < width - && !visited.contains(&(x, y + 1)) - && Solution::dfs(x, y + 1, &seq[1..], board, visited.clone(), height, width)); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_79() { - assert_eq!(Solution::exist(vec![vec!['a']], "a".to_owned()), true); - assert_eq!( - Solution::exist( - vec![ - vec!['A', 'B', 'C', 'E'], - vec!['S', 'F', 'C', 'S'], - vec!['A', 'D', 'E', 'E'], - ], - "ABCCED".to_owned() - ), - true - ); - assert_eq!( - Solution::exist( - vec![ - vec!['A', 'B', 'C', 'E'], - vec!['S', 'F', 'C', 'S'], - vec!['A', 'D', 'E', 'E'], - ], - "SEE".to_owned() - ), - true - ); - assert_eq!( - Solution::exist( - vec![ - vec!['A', 'B', 'C', 'E'], - vec!['S', 'F', 'C', 'S'], - vec!['A', 'D', 'E', 'E'], - ], - "ABCB".to_owned() - ), - false - ); - } -} diff --git a/src/solution/s0080_remove_duplicates_from_sorted_array_ii.rs b/src/solution/s0080_remove_duplicates_from_sorted_array_ii.rs deleted file mode 100644 index 70b3cf82..00000000 --- a/src/solution/s0080_remove_duplicates_from_sorted_array_ii.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [80] Remove Duplicates from Sorted Array II - * - * Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. - * - * Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. - * - * Example 1: - * - * - * Given nums = [1,1,1,2,2,3], - * - * Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. - * - * It doesn't matter what you leave beyond the returned length. - * - * Example 2: - * - * - * Given nums = [0,0,1,1,1,1,2,3,3], - * - * Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively. - * - * It doesn't matter what values are set beyond the returned length. - * - * - * Clarification: - * - * Confused why the returned value is an integer but your answer is an array? - * - * Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. - * - * Internally you can think of this: - * - * - * // nums is passed in by reference. (i.e., without making a copy) - * int len = removeDuplicates(nums); - * - * // any modification to nums in your function would be known by the caller. - * // using the length returned by your function, it prints the first len elements. - * for (int i = 0; i < len; i++) { - * print(nums[i]); - * } - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/ -// discuss: https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn remove_duplicates(nums: &mut Vec) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_80() {} -} diff --git a/src/solution/s0081_search_in_rotated_sorted_array_ii.rs b/src/solution/s0081_search_in_rotated_sorted_array_ii.rs deleted file mode 100644 index 7972a16f..00000000 --- a/src/solution/s0081_search_in_rotated_sorted_array_ii.rs +++ /dev/null @@ -1,52 +0,0 @@ -/** - * [81] Search in Rotated Sorted Array II - * - * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. - * - * (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). - * - * You are given a target value to search. If found in the array return true, otherwise return false. - * - * Example 1: - * - * - * Input: nums = [2,5,6,0,0,1,2], target = 0 - * Output: true - * - * - * Example 2: - * - * - * Input: nums = [2,5,6,0,0,1,2], target = 3 - * Output: false - * - * Follow up: - * - * - * This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates. - * Would this affect the run-time complexity? How and why? - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ -// discuss: https://leetcode.com/problems/search-in-rotated-sorted-array-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn search(nums: Vec, target: i32) -> bool { - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_81() {} -} diff --git a/src/solution/s0082_remove_duplicates_from_sorted_list_ii.rs b/src/solution/s0082_remove_duplicates_from_sorted_list_ii.rs deleted file mode 100644 index 067d0c79..00000000 --- a/src/solution/s0082_remove_duplicates_from_sorted_list_ii.rs +++ /dev/null @@ -1,59 +0,0 @@ -/** - * [82] Remove Duplicates from Sorted List II - * - * Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. - * - * Example 1: - * - * - * Input: 1->2->3->3->4->4->5 - * Output: 1->2->5 - * - * - * Example 2: - * - * - * Input: 1->1->1->2->3 - * Output: 2->3 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/ -// discuss: https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for singly-linked list. -// #[derive(PartialEq, Eq, Debug)] -// pub struct ListNode { -// pub val: i32, -// pub next: Option> -// } -// -// impl ListNode { -// #[inline] -// fn new(val: i32) -> Self { -// ListNode { -// next: None, -// val -// } -// } -// } -impl Solution { - pub fn delete_duplicates(head: Option>) -> Option> { - None - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_82() {} -} diff --git a/src/solution/s0083_remove_duplicates_from_sorted_list.rs b/src/solution/s0083_remove_duplicates_from_sorted_list.rs deleted file mode 100644 index 5c9552cc..00000000 --- a/src/solution/s0083_remove_duplicates_from_sorted_list.rs +++ /dev/null @@ -1,59 +0,0 @@ -/** - * [83] Remove Duplicates from Sorted List - * - * Given a sorted linked list, delete all duplicates such that each element appear only once. - * - * Example 1: - * - * - * Input: 1->1->2 - * Output: 1->2 - * - * - * Example 2: - * - * - * Input: 1->1->2->3->3 - * Output: 1->2->3 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/remove-duplicates-from-sorted-list/ -// discuss: https://leetcode.com/problems/remove-duplicates-from-sorted-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for singly-linked list. -// #[derive(PartialEq, Eq, Debug)] -// pub struct ListNode { -// pub val: i32, -// pub next: Option> -// } -// -// impl ListNode { -// #[inline] -// fn new(val: i32) -> Self { -// ListNode { -// next: None, -// val -// } -// } -// } -impl Solution { - pub fn delete_duplicates(head: Option>) -> Option> { - None - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_83() {} -} diff --git a/src/solution/s0084_largest_rectangle_in_histogram.rs b/src/solution/s0084_largest_rectangle_in_histogram.rs deleted file mode 100644 index d33b9bda..00000000 --- a/src/solution/s0084_largest_rectangle_in_histogram.rs +++ /dev/null @@ -1,91 +0,0 @@ -/** - * [84] Largest Rectangle in Histogram - * - * Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram. - * - * - * - *
- * Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]. - * - * - * - *
- * The largest rectangle is shown in the shaded area, which has area = 10 unit. - * - * - * - * Example: - * - * - * Input: [2,1,5,6,2,3] - * Output: 10 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/largest-rectangle-in-histogram/ -// discuss: https://leetcode.com/problems/largest-rectangle-in-histogram/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// record the height and start position using 2 stack, thus we reuse the previously scanned information -impl Solution { - pub fn largest_rectangle_area(heights: Vec) -> i32 { - let mut positions = Vec::new(); - let mut hs = Vec::new(); - let mut max_area = 0; - let len = heights.len(); - for (i, h) in heights.into_iter().enumerate() { - let mut last_pop = None; - while hs.last().is_some() && *hs.last().unwrap() >= h { - max_area = i32::max( - max_area, - hs.last().unwrap() * ((i - positions.last().unwrap()) as i32), - ); - hs.pop(); - last_pop = positions.pop(); - } - if last_pop.is_some() { - positions.push(last_pop.unwrap()); - } else { - positions.push(i); - } - hs.push(h); - } - // drain stack - while !hs.is_empty() { - max_area = i32::max( - max_area, - hs.last().unwrap() * ((len - positions.last().unwrap()) as i32), - ); - positions.pop(); - hs.pop(); - } - max_area - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_84() { - assert_eq!(Solution::largest_rectangle_area(vec![2, 1, 5, 6, 2, 3]), 10); - assert_eq!( - Solution::largest_rectangle_area(vec![1, 1, 1, 1, 1, 1, 1, 1]), - 8 - ); - assert_eq!(Solution::largest_rectangle_area(vec![2, 2]), 4); - assert_eq!( - Solution::largest_rectangle_area(vec![1, 2, 8, 8, 2, 2, 1]), - 16 - ); - assert_eq!(Solution::largest_rectangle_area(vec![2, 1, 2]), 3); - assert_eq!(Solution::largest_rectangle_area(vec![1, 3, 2, 1, 2]), 5); - } -} diff --git a/src/solution/s0085_maximal_rectangle.rs b/src/solution/s0085_maximal_rectangle.rs deleted file mode 100644 index 6dbf2951..00000000 --- a/src/solution/s0085_maximal_rectangle.rs +++ /dev/null @@ -1,43 +0,0 @@ -/** - * [85] Maximal Rectangle - * - * Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area. - * - * Example: - * - * - * Input: - * [ - * ["1","0","1","0","0"], - * ["1","0","1","1","1"], - * ["1","1","1","1","1"], - * ["1","0","0","1","0"] - * ] - * Output: 6 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/maximal-rectangle/ -// discuss: https://leetcode.com/problems/maximal-rectangle/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn maximal_rectangle(matrix: Vec>) -> i32 { - let mut max_area = 0; - - max_area - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_85() {} -} diff --git a/src/solution/s0086_partition_list.rs b/src/solution/s0086_partition_list.rs deleted file mode 100644 index 1f4952dd..00000000 --- a/src/solution/s0086_partition_list.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [86] Partition List - * - * Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. - * - * You should preserve the original relative order of the nodes in each of the two partitions. - * - * Example: - * - * - * Input: head = 1->4->3->2->5->2, x = 3 - * Output: 1->2->2->4->3->5 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/partition-list/ -// discuss: https://leetcode.com/problems/partition-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn partition(head: Option>, x: i32) -> Option> { - let mut lower = Some(Box::new(ListNode::new(0))); - let mut higher = Some(Box::new(ListNode::new(0))); - let mut lower_tail = lower.as_mut(); - let mut higher_tail = higher.as_mut(); - let mut head = head; - while let Some(mut inner) = head { - let mut next = inner.next.take(); - if inner.val < x { - lower_tail.as_mut().unwrap().next = Some(inner); - lower_tail = lower_tail.unwrap().next.as_mut(); - } else { - higher_tail.as_mut().unwrap().next = Some(inner); - higher_tail = higher_tail.unwrap().next.as_mut(); - } - head = next - } - lower_tail.as_mut().unwrap().next = higher.unwrap().next.take(); - lower.unwrap().next.take() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_86() { - assert_eq!( - Solution::partition(linked![1, 4, 3, 2, 5, 2], 3), - linked![1, 2, 2, 4, 3, 5] - ); - assert_eq!( - Solution::partition(linked![1, 4, 3, 2, 5, 2], 8), - linked![1, 4, 3, 2, 5, 2] - ); - assert_eq!(Solution::partition(linked![], 0), linked![]); - } -} diff --git a/src/solution/s0087_scramble_string.rs b/src/solution/s0087_scramble_string.rs deleted file mode 100644 index 19b47dc0..00000000 --- a/src/solution/s0087_scramble_string.rs +++ /dev/null @@ -1,85 +0,0 @@ -/** - * [87] Scramble String - * - * Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. - * - * Below is one possible representation of s1 = "great": - * - * - * great - * / \ - * gr eat - * / \ / \ - * g r e at - * / \ - * a t - * - * - * To scramble the string, we may choose any non-leaf node and swap its two children. - * - * For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat". - * - * - * rgeat - * / \ - * rg eat - * / \ / \ - * r g e at - * / \ - * a t - * - * - * We say that "rgeat" is a scrambled string of "great". - * - * Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae". - * - * - * rgtae - * / \ - * rg tae - * / \ / \ - * r g ta e - * / \ - * t a - * - * - * We say that "rgtae" is a scrambled string of "great". - * - * Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1. - * - * Example 1: - * - * - * Input: s1 = "great", s2 = "rgeat" - * Output: true - * - * - * Example 2: - * - * - * Input: s1 = "abcde", s2 = "caebd" - * Output: false - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/scramble-string/ -// discuss: https://leetcode.com/problems/scramble-string/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_scramble(s1: String, s2: String) -> bool { - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_87() {} -} diff --git a/src/solution/s0088_merge_sorted_array.rs b/src/solution/s0088_merge_sorted_array.rs deleted file mode 100644 index bdcc2a1a..00000000 --- a/src/solution/s0088_merge_sorted_array.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [88] Merge Sorted Array - * - * Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. - * - * Note: - * - * - * The number of elements initialized in nums1 and nums2 are m and n respectively. - * You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. - * - * - * Example: - * - * - * Input: - * nums1 = [1,2,3,0,0,0], m = 3 - * nums2 = [2,5,6], n = 3 - * - * Output: [1,2,2,3,5,6] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/merge-sorted-array/ -// discuss: https://leetcode.com/problems/merge-sorted-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn merge(nums1: &mut Vec, m: i32, nums2: &mut Vec, n: i32) { - let mut i = m - 1; - let mut j = n - 1; - let mut z = m + n - 1; - while z >= 0 { - if i >= 0 && (j < 0 || nums1[i as usize] >= nums2[j as usize]) { - nums1[z as usize] = nums1[i as usize]; - i -= 1; - } else { - nums1[z as usize] = nums2[j as usize]; - j -= 1; - } - z -= 1; - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_88() { - let mut vec1 = vec![1, 2, 3, 0, 0, 0]; - let mut vec2 = vec![2, 5, 6]; - Solution::merge(&mut vec1, 3, &mut vec2, 3); - assert_eq!(vec1, vec![1, 2, 2, 3, 5, 6]); - - let mut vec1 = vec![1, 2, 3]; - let mut vec2 = vec![]; - Solution::merge(&mut vec1, 3, &mut vec2, 0); - assert_eq!(vec1, vec![1, 2, 3]); - - let mut vec1 = vec![0, 0, 0]; - let mut vec2 = vec![1, 2, 3]; - Solution::merge(&mut vec1, 0, &mut vec2, 3); - assert_eq!(vec1, vec![1, 2, 3]); - } -} diff --git a/src/solution/s0089_gray_code.rs b/src/solution/s0089_gray_code.rs deleted file mode 100644 index 039283eb..00000000 --- a/src/solution/s0089_gray_code.rs +++ /dev/null @@ -1,82 +0,0 @@ -/** - * [89] Gray Code - * - * The gray code is a binary numeral system where two successive values differ in only one bit. - * - * Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0. - * - * Example 1: - * - * - * Input: 2 - * Output: [0,1,3,2] - * Explanation: - * 00 - 0 - * 01 - 1 - * 11 - 3 - * 10 - 2 - * - * For a given n, a gray code sequence may not be uniquely defined. - * For example, [0,2,3,1] is also a valid gray code sequence. - * - * 00 - 0 - * 10 - 2 - * 11 - 3 - * 01 - 1 - * - * - * Example 2: - * - * - * Input: 0 - * Output: [0] - * Explanation: We define the gray code sequence to begin with 0. - * A gray code sequence of n has size = 2^n, which for n = 0 the size is 2^0 = 1. - * Therefore, for n = 0 the gray code sequence is [0]. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/gray-code/ -// discuss: https://leetcode.com/problems/gray-code/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -0000 -0001 <- flip [0] to 1, traverse [] in reverse order -0011 <- flip [1] to 1, traverse [0] in reverse order -0010 -0110 <- flip [2] to 1, traverse [1,0] in reverse order -0111 -0101 -0100 -1100 <- flip [3] to 1, traverse [2,1,0] in reverse order - */ -impl Solution { - pub fn gray_code(n: i32) -> Vec { - let mut res = vec![0]; - for i in 0..n { - for j in (0..res.len()).rev() { - res.push(2_i32.pow(i as u32) + res[j]); - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_89() { - assert_eq!(Solution::gray_code(2), vec![0, 1, 3, 2]); - assert_eq!(Solution::gray_code(1), vec![0, 1]); - assert_eq!(Solution::gray_code(0), vec![0]); - assert_eq!(Solution::gray_code(3), vec![0, 1, 3, 2, 6, 7, 5, 4]); - } -} diff --git a/src/solution/s0090_subsets_ii.rs b/src/solution/s0090_subsets_ii.rs deleted file mode 100644 index e859e2d1..00000000 --- a/src/solution/s0090_subsets_ii.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [90] Subsets II - * - * Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set). - * - * Note: The solution set must not contain duplicate subsets. - * - * Example: - * - * - * Input: [1,2,2] - * Output: - * [ - * [2], - * [1], - * [1,2,2], - * [2,2], - * [1,2], - * [] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/subsets-ii/ -// discuss: https://leetcode.com/problems/subsets-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -count the repeats of each number, -then in backtracking, each number can be picked up for 0..repeat times - -using BTreeMap to preserve order (easy for test) -*/ -use std::collections::BTreeMap; -impl Solution { - pub fn subsets_with_dup(nums: Vec) -> Vec> { - let mut res = Vec::new(); - let nums = nums - .into_iter() - .fold(BTreeMap::new(), |mut map, v| { - *map.entry(v).or_insert(0) += 1; - map - }) - .into_iter() - .collect::>(); - Solution::backtrack(0, vec![], &nums, &mut res); - res - } - - fn backtrack( - start: usize, - mut curr: Vec, - nums: &Vec<(i32, i32)>, - result: &mut Vec>, - ) { - if start >= nums.len() { - result.push(curr); - return; - } - for repeat in 0..nums[start].1 + 1 { - let mut inner = curr.clone(); - for _ in 0..repeat { - inner.push(nums[start].0); - } - Solution::backtrack(start + 1, inner, nums, result); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_90() { - assert_eq!( - Solution::subsets_with_dup(vec![1, 2, 2]), - vec![ - vec![], - vec![2], - vec![2, 2], - vec![1], - vec![1, 2], - vec![1, 2, 2], - ] - ); - assert_eq!(Solution::subsets_with_dup(vec![1]), vec![vec![], vec![1],]); - assert_eq!(Solution::subsets_with_dup(vec![]), vec![vec![],]); - } -} diff --git a/src/solution/s0091_decode_ways.rs b/src/solution/s0091_decode_ways.rs deleted file mode 100644 index bbecc932..00000000 --- a/src/solution/s0091_decode_ways.rs +++ /dev/null @@ -1,52 +0,0 @@ -/** - * [91] Decode Ways - * - * A message containing letters from A-Z is being encoded to numbers using the following mapping: - * - * - * 'A' -> 1 - * 'B' -> 2 - * ... - * 'Z' -> 26 - * - * - * Given a non-empty string containing only digits, determine the total number of ways to decode it. - * - * Example 1: - * - * - * Input: "12" - * Output: 2 - * Explanation: It could be decoded as "AB" (1 2) or "L" (12). - * - * - * Example 2: - * - * - * Input: "226" - * Output: 3 - * Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6). - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/decode-ways/ -// discuss: https://leetcode.com/problems/decode-ways/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn num_decodings(s: String) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_91() {} -} diff --git a/src/solution/s0092_reverse_linked_list_ii.rs b/src/solution/s0092_reverse_linked_list_ii.rs deleted file mode 100644 index 318c0d15..00000000 --- a/src/solution/s0092_reverse_linked_list_ii.rs +++ /dev/null @@ -1,54 +0,0 @@ -/** - * [92] Reverse Linked List II - * - * Reverse a linked list from position m to n. Do it in one-pass. - * - * Note: 1 ≤ m ≤ n ≤ length of list. - * - * Example: - * - * - * Input: 1->2->3->4->5->NULL, m = 2, n = 4 - * Output: 1->4->3->2->5->NULL - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/reverse-linked-list-ii/ -// discuss: https://leetcode.com/problems/reverse-linked-list-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for singly-linked list. -// #[derive(PartialEq, Eq, Debug)] -// pub struct ListNode { -// pub val: i32, -// pub next: Option> -// } -// -// impl ListNode { -// #[inline] -// fn new(val: i32) -> Self { -// ListNode { -// next: None, -// val -// } -// } -// } -impl Solution { - pub fn reverse_between(head: Option>, m: i32, n: i32) -> Option> { - None - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_92() {} -} diff --git a/src/solution/s0093_restore_ip_addresses.rs b/src/solution/s0093_restore_ip_addresses.rs deleted file mode 100644 index 5c2706b4..00000000 --- a/src/solution/s0093_restore_ip_addresses.rs +++ /dev/null @@ -1,35 +0,0 @@ -/** - * [93] Restore IP Addresses - * - * Given a string containing only digits, restore it by returning all possible valid IP address combinations. - * - * Example: - * - * - * Input: "25525511135" - * Output: ["255.255.11.135", "255.255.111.35"] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/restore-ip-addresses/ -// discuss: https://leetcode.com/problems/restore-ip-addresses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn restore_ip_addresses(s: String) -> Vec { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_93() {} -} diff --git a/src/solution/s0094_binary_tree_inorder_traversal.rs b/src/solution/s0094_binary_tree_inorder_traversal.rs deleted file mode 100644 index 01afb57a..00000000 --- a/src/solution/s0094_binary_tree_inorder_traversal.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [94] Binary Tree Inorder Traversal - * - * Given a binary tree, return the inorder traversal of its nodes' values. - * - * Example: - * - * - * Input: [1,null,2,3] - * 1 - * \ - * 2 - * / - * 3 - * - * Output: [1,3,2] - * - * Follow up: Recursive solution is trivial, could you do it iteratively? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/binary-tree-inorder-traversal/ -// discuss: https://leetcode.com/problems/binary-tree-inorder-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use crate::util::tree::{to_tree, TreeNode}; -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn inorder_traversal(root: Option>>) -> Vec { - let mut res = Vec::new(); - Solution::inorder_traverse(root.as_ref(), &mut (|v| res.push(v))); - res - } - - fn inorder_traverse(root: Option<&Rc>>, consumer: &mut F) { - if let Some(node) = root { - Solution::inorder_traverse(node.borrow().left.as_ref(), consumer); - consumer(root.as_ref().unwrap().borrow().val); - Solution::inorder_traverse(node.borrow().right.as_ref(), consumer); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_94() { - assert_eq!( - Solution::inorder_traversal(tree![1, null, 2, 3]), - vec![1, 3, 2] - ); - assert_eq!( - Solution::inorder_traversal(tree![1, 2, 3, 4, 5, 6, 7]), - vec![4, 2, 5, 1, 6, 3, 7] - ); - } -} diff --git a/src/solution/s0095_unique_binary_search_trees_ii.rs b/src/solution/s0095_unique_binary_search_trees_ii.rs deleted file mode 100644 index 1f75639d..00000000 --- a/src/solution/s0095_unique_binary_search_trees_ii.rs +++ /dev/null @@ -1,89 +0,0 @@ -/** - * [95] Unique Binary Search Trees II - * - * Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n. - * - * Example: - * - * - * Input: 3 - * Output: - * [ - * [1,null,3,2], - * [3,2,null,1], - * [3,1,null,null,2], - * [2,1,3], - * [1,null,2,null,3] - * ] - * Explanation: - * The above output corresponds to the 5 unique BST's shown below: - * - * 1 3 3 2 1 - * \ / / / \ \ - * 3 2 1 1 3 2 - * / / \ \ - * 2 1 2 3 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/unique-binary-search-trees-ii/ -// discuss: https://leetcode.com/problems/unique-binary-search-trees-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* - 1 - - 1 2 - \ / - 2 1 - - 1 3 3 2 1 - \ / / / \ \ - 3 2 1 1 3 2 - / / \ \ - 2 1 2 3 - - 4 1 - / \ -(5*) 3 - / \ - 2 4 - */ -use crate::util::tree::{to_tree, TreeNode}; -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn generate_trees(n: i32) -> Vec>>> { - if n < 1 { - return vec![]; - } - let mut res = vec![Some(Rc::new(RefCell::new(TreeNode::new(1))))]; - for val in 2..n + 1 { - let mut next = Vec::new(); - for root in res.into_iter() { - let mut dummy = Some(Rc::new(RefCell::new(TreeNode { - val: 0, - left: None, - right: None, - }))); - let mut parent = dummy.as_ref().unwrap().clone(); - let mut node = root; - // we know that val is larger than all the elements in the tree - } - res = next; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_95() {} -} diff --git a/src/solution/s0096_unique_binary_search_trees.rs b/src/solution/s0096_unique_binary_search_trees.rs deleted file mode 100644 index 4f64c26c..00000000 --- a/src/solution/s0096_unique_binary_search_trees.rs +++ /dev/null @@ -1,43 +0,0 @@ -/** - * [96] Unique Binary Search Trees - * - * Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n? - * - * Example: - * - * - * Input: 3 - * Output: 5 - * Explanation: - * Given n = 3, there are a total of 5 unique BST's: - * - * 1 3 3 2 1 - * \ / / / \ \ - * 3 2 1 1 3 2 - * / / \ \ - * 2 1 2 3 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/unique-binary-search-trees/ -// discuss: https://leetcode.com/problems/unique-binary-search-trees/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn num_trees(n: i32) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_96() {} -} diff --git a/src/solution/s0097_interleaving_string.rs b/src/solution/s0097_interleaving_string.rs deleted file mode 100644 index 8d69d851..00000000 --- a/src/solution/s0097_interleaving_string.rs +++ /dev/null @@ -1,115 +0,0 @@ -/** - * [97] Interleaving String - * - * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. - * - * Example 1: - * - * - * Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac" - * Output: true - * - * - * Example 2: - * - * - * Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc" - * Output: false - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/interleaving-string/ -// discuss: https://leetcode.com/problems/interleaving-string/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// DFS with memorization -/* -思路: DFS, 三个指针 i,j,k 分别指向 s1, s2, s3 已经消费到的 char 位置, 下一个可以走的路径是 s3 当前消费到的 char 值 - -如 aaaaaas aaaaaaaw aaaaaaaaaaaaaasw -那么第一步可以从 s1 或 s2 取一个 char, 用 DFS 的方式搜索整个解空间 - -优化: 直接 DFS 非常慢, 还是上面的例子, 最差情况是大量重复字符, 时间复杂度直接是 2^(M+N), 优化方式借鉴 DP 经常用到的 -memorize, 使用一个二维数组缓存每一对遍历过的 i,j 最后是否能产生合法的 interleaving. - -优化后通过缓存剪除的路径比较难分析, 但很显然能知道最差情况也只需要将所有 M*N 的组合进行标记, 因此最差时间复杂度 O(M*N) -空间复杂度 O(M*N) -*/ - -impl Solution { - pub fn is_interleave(s1: String, s2: String, s3: String) -> bool { - let mut cache = vec![vec![false; s2.len() + 1]; s1.len() + 1]; - Solution::dfs( - &s1.chars().collect(), - &s2.chars().collect(), - &s3.chars().collect(), - 0, - 0, - 0, - &mut cache, - ) - } - - fn dfs( - s1: &Vec, - s2: &Vec, - s3: &Vec, - i: usize, - j: usize, - k: usize, - invalid: &mut Vec>, - ) -> bool { - if invalid[i][j] { - return false; - } - if i == s1.len() && j == s2.len() && k == s3.len() { - return true; - } - let valid = (i < s1.len() - && k < s3.len() - && s1[i] == s3[k] - && Solution::dfs(s1, s2, s3, i + 1, j, k + 1, invalid)) - || (j < s2.len() - && k < s3.len() - && s2[j] == s3[k] - && Solution::dfs(s1, s2, s3, i, j + 1, k + 1, invalid)); - if !valid { - invalid[i][j] = true - } - valid - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_97() { - assert_eq!( - Solution::is_interleave( - "aabcc".to_owned(), - "dbbca".to_owned(), - "aadbbcbcac".to_owned() - ), - true - ); - assert_eq!( - Solution::is_interleave( - "aabcc".to_owned(), - "dbbca".to_owned(), - "aadbbbaccc".to_owned() - ), - false - ); - assert_eq!( - Solution::is_interleave("a".to_owned(), "b".to_owned(), "a".to_owned()), - false - ); - } -} diff --git a/src/solution/s0098_validate_binary_search_tree.rs b/src/solution/s0098_validate_binary_search_tree.rs deleted file mode 100644 index 09098c18..00000000 --- a/src/solution/s0098_validate_binary_search_tree.rs +++ /dev/null @@ -1,97 +0,0 @@ -/** - * [98] Validate Binary Search Tree - * - * Given a binary tree, determine if it is a valid binary search tree (BST). - * - * Assume a BST is defined as follows: - * - * - * The left subtree of a node contains only nodes with keys less than the node's key. - * The right subtree of a node contains only nodes with keys greater than the node's key. - * Both the left and right subtrees must also be binary search trees. - * - * - * Example 1: - * - * - * Input: - * 2 - * / \ - * 1 3 - * Output: true - * - * - * Example 2: - * - * - * 5 - * / \ - * 1 4 - * / \ - * 3 6 - * Output: false - * Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value - * is 5 but its right child's value is 4. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/validate-binary-search-tree/ -// discuss: https://leetcode.com/problems/validate-binary-search-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for a binary tree node. -use crate::util::tree::{to_tree, TreeNode}; - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn is_valid_bst(root: Option>>) -> bool { - let mut vec = vec![]; - Solution::preorder_traverse(root.as_ref(), &mut vec) - } - - fn preorder_traverse( - root: Option<&Rc>>, - formers: &mut Vec<(i32, i32)>, - ) -> bool { - if let Some(node) = root { - let root_val = root.as_ref().unwrap().borrow().val; - for former in formers.iter() { - if (former.0 < 0 && root_val >= former.1) || (former.0 > 0 && root_val <= former.1) - { - return false; - } - } - let mut to_right = formers.clone(); - formers.push((-1, root_val)); - to_right.push((1, root_val)); - Solution::preorder_traverse(node.borrow().left.as_ref(), formers) - && Solution::preorder_traverse(node.borrow().right.as_ref(), &mut to_right) - } else { - true - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_98() { - assert_eq!( - Solution::is_valid_bst(tree![5, 1, 4, null, null, 3, 6]), - false - ); - assert_eq!(Solution::is_valid_bst(tree![2, 1, 3]), true); - assert_eq!( - Solution::is_valid_bst(tree![10, 5, 15, null, null, 6, 20]), - false - ); - } -} diff --git a/src/solution/s0099_recover_binary_search_tree.rs b/src/solution/s0099_recover_binary_search_tree.rs deleted file mode 100644 index 2401231a..00000000 --- a/src/solution/s0099_recover_binary_search_tree.rs +++ /dev/null @@ -1,162 +0,0 @@ -/** - * [99] Recover Binary Search Tree - * - * Two elements of a binary search tree (BST) are swapped by mistake. - * - * Recover the tree without changing its structure. - * - * Example 1: - * - * - * Input: [1,3,null,null,2] - * - * 1 - * / - * 3 - * \ - * 2 - * - * Output: [3,1,null,null,2] - * - * 3 - * / - * 1 - * \ - * 2 - * - * - * Example 2: - * - * - * Input: [3,1,4,null,null,2] - * - * 3 - * / \ - * 1 4 - * / - * 2 - * - * Output: [2,1,4,null,null,3] - * - * 2 - * / \ - * 1 4 - * / - * 3 - * - * - * Follow up: - * - * - * A solution using O(n) space is pretty straight forward. - * Could you devise a constant space solution? - * - * - */ -pub struct Solution {} - -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/recover-binary-search-tree/ -// discuss: https://leetcode.com/problems/recover-binary-search-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -基本思路: 对 root, 收集左右两侧的节点值: - - - 右侧有值小于 root, 左侧有值大于 root: 交换这两个值 - - 一侧有值不合法, 另一侧合法: 交换不合法的值与 root - - 两侧都合法: 递归对左右子树进行上述操作 - -直接使用上述思路会重复迭代子树, 因此可以先用一次中序遍历, 将值的分布记录在数组中, 之后可以用 O(1) 的时间得到某个 node 两侧的值分布 - -这种办法的时间空间复杂度都是 O(N) - -原题中的 follow up 部分提出可以想一个常数空间复杂度的算法, 因此可以继续优化. - -最开始的办法会重复迭代子树是因为 Top-down, 可以尝试一下 Bottom-up 能不能解决问题: - -- 后序遍历递归校验每个 node 是否合法 -- 假如 node 合法, 那么只需要把这个子树中的最大值和最小值返回到上一个层级来帮助判断上一个层级的子树是否合法即可, 无需记录这个子树的所有值 -- 假如 node 不合法, 按最开始的办法进行处理, 除非... -- 只有一侧值不合法, 这时候由于是 Bottom-up, 不能直接交换不合法的值与 root, 而要判断交换后能否合法(Top-down 的办法中由于题目给定了有且仅有 swap 一次, 因此与 root 交换必然是合法的) - -这个办法时间复杂度 O(N), 空间 O(1). 题解就用 Bottom-up 来写. - */ -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn recover_tree(root: &mut Option>>) { - Solution::recover_helper(root.as_ref()); - } - - fn recover_helper( - root: Option<&Rc>>, - ) -> ( - Option>>, - Option>>, - bool, - ) { - if let Some(node) = root { - let (l_min, l_max, l_flag) = Solution::recover_helper(node.borrow().left.as_ref()); - let (r_min, r_max, r_flag) = Solution::recover_helper(node.borrow().right.as_ref()); - // we've already find a swap, return quickly - if l_flag || r_flag { - return (None, None, true); - } - let root_val = node.borrow().val; - let l_err = l_max.as_ref().map_or(false, |v| v.borrow().val > root_val); - let r_err = r_min.as_ref().map_or(false, |v| v.borrow().val < root_val); - // invalid sub-tree found, do swap - if l_err || r_err { - if l_err && r_err { - } else if l_err { - std::mem::swap( - &mut l_max.unwrap().borrow_mut().val, - &mut node.borrow_mut().val, - ); - } else if r_err { - std::mem::swap( - &mut r_min.unwrap().borrow_mut().val, - &mut node.borrow_mut().val, - ) - } - return (None, None, true); - } - ( - if l_min.is_some() { - l_min - } else { - Some(node.clone()) - }, - if r_max.is_some() { - r_max - } else { - Some(node.clone()) - }, - false, - ) - } else { - (None, None, false) - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_99() { - let mut tree = tree![3, 1, 4, null, null, 2]; - Solution::recover_tree(&mut tree); - assert_eq!(tree, tree![2, 1, 4, null, null, 3]); - - let mut tree = tree![2, 6, 5, null, null, 3, 1, null, 4]; - Solution::recover_tree(&mut tree); - assert_eq!(tree, tree![2, 1, 5, null, null, 3, 6, null, 4]); - } -} diff --git a/src/solution/s0100_same_tree.rs b/src/solution/s0100_same_tree.rs deleted file mode 100644 index c7dc1372..00000000 --- a/src/solution/s0100_same_tree.rs +++ /dev/null @@ -1,76 +0,0 @@ -/** - * [100] Same Tree - * - * Given two binary trees, write a function to check if they are the same or not. - * - * Two binary trees are considered the same if they are structurally identical and the nodes have the same value. - * - * Example 1: - * - * - * Input: 1 1 - * / \ / \ - * 2 3 2 3 - * - * [1,2,3], [1,2,3] - * - * Output: true - * - * - * Example 2: - * - * - * Input: 1 1 - * / \ - * 2 2 - * - * [1,2], [1,null,2] - * - * Output: false - * - * - * Example 3: - * - * - * Input: 1 1 - * / \ / \ - * 2 1 1 2 - * - * [1,2,1], [1,1,2] - * - * Output: false - * - * - */ -pub struct Solution {} - -use crate::util::tree::{to_tree, TreeNode}; -// problem: https://leetcode.com/problems/same-tree/ -// discuss: https://leetcode.com/problems/same-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn is_same_tree( - p: Option>>, - q: Option>>, - ) -> bool { - p == q - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_100() { - assert_eq!( - Solution::is_same_tree(tree![1, 2, 3, 4, null, 5], tree![1, 2, 3, 4, null, 5]), - true - ) - } -} diff --git a/src/solution/s0101_symmetric_tree.rs b/src/solution/s0101_symmetric_tree.rs deleted file mode 100644 index 41f44bc8..00000000 --- a/src/solution/s0101_symmetric_tree.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [101] Symmetric Tree - * - * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). - * - * - * For example, this binary tree [1,2,2,3,4,4,3] is symmetric: - * - * 1 - * / \ - * 2 2 - * / \ / \ - * 3 4 4 3 - * - * - * - * But the following [1,2,2,null,3,null,3] is not:
- * - * 1 - * / \ - * 2 2 - * \ \ - * 3 3 - * - * - * - * - * Note:
- * Bonus points if you could solve it both recursively and iteratively. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/symmetric-tree/ -// discuss: https://leetcode.com/problems/symmetric-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn is_symmetric(root: Option>>) -> bool { - Solution::symmetric_helper( - root.as_ref().and_then(|v| v.borrow().left.clone()), - root.as_ref().and_then(|v| v.borrow().right.clone()), - ) - } - - fn symmetric_helper( - left: Option>>, - right: Option>>, - ) -> bool { - match (left, right) { - (Some(left), Some(right)) => { - left.borrow().val == right.borrow().val - && Solution::symmetric_helper( - left.borrow().left.clone(), - right.borrow().right.clone(), - ) - && Solution::symmetric_helper( - left.borrow().right.clone(), - right.borrow().left.clone(), - ) - } - (None, None) => true, - _ => false, - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_101() { - assert_eq!(Solution::is_symmetric(tree![1, 2, 2, 3, 4, 4, 3]), true); - assert_eq!( - Solution::is_symmetric(tree![1, 2, 2, null, 3, null, 3]), - false - ); - assert_eq!(Solution::is_symmetric(tree![]), true); - } -} diff --git a/src/solution/s0102_binary_tree_level_order_traversal.rs b/src/solution/s0102_binary_tree_level_order_traversal.rs deleted file mode 100644 index a5a3fd13..00000000 --- a/src/solution/s0102_binary_tree_level_order_traversal.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [102] Binary Tree Level Order Traversal - * - * Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). - * - * - * For example:
- * Given binary tree [3,9,20,null,null,15,7],
- * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * - * - * return its level order traversal as:
- * - * [ - * [3], - * [9,20], - * [15,7] - * ] - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-level-order-traversal/ -// discuss: https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn level_order(root: Option>>) -> Vec> { - let mut res = Vec::new(); - let mut current_level = 0; - if root.is_none() { - return res; - } - let mut deq = VecDeque::new(); - deq.push_back((0, root.clone())); - let mut vec = Vec::new(); - while !deq.is_empty() { - if let Some((level, Some(node))) = deq.pop_front() { - deq.push_back((level + 1, node.borrow().left.clone())); - deq.push_back((level + 1, node.borrow().right.clone())); - if level > current_level { - res.push(vec); - vec = Vec::new(); - current_level = level; - } - vec.push(node.borrow().val); - } - } - if !vec.is_empty() { - res.push(vec) - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_102() { - assert_eq!( - Solution::level_order(tree![3, 9, 20, null, null, 15, 7]), - vec![vec![3], vec![9, 20], vec![15, 7]] - ); - } -} diff --git a/src/solution/s0103_binary_tree_zigzag_level_order_traversal.rs b/src/solution/s0103_binary_tree_zigzag_level_order_traversal.rs deleted file mode 100644 index 78aa7b5e..00000000 --- a/src/solution/s0103_binary_tree_zigzag_level_order_traversal.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [103] Binary Tree Zigzag Level Order Traversal - * - * Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). - * - * - * For example:
- * Given binary tree [3,9,20,null,null,15,7],
- * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * - * - * return its zigzag level order traversal as:
- * - * [ - * [3], - * [20,9], - * [15,7] - * ] - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ -// discuss: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn zigzag_level_order(root: Option>>) -> Vec> { - let mut res = Vec::new(); - let mut current_level = 0; - if root.is_none() { - return res; - } - let mut deq = VecDeque::new(); - deq.push_back((0, root.clone())); - let mut vec = Vec::new(); - while !deq.is_empty() { - if let Some((level, Some(node))) = deq.pop_front() { - deq.push_back((level + 1, node.borrow().left.clone())); - deq.push_back((level + 1, node.borrow().right.clone())); - if level > current_level { - if current_level % 2 == 1 { - vec.reverse(); - } - res.push(vec); - vec = Vec::new(); - current_level = level; - } - vec.push(node.borrow().val); - } - } - if !vec.is_empty() { - if current_level % 2 == 1 { - vec.reverse(); - } - res.push(vec) - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_103() { - assert_eq!( - Solution::zigzag_level_order(tree![3, 9, 20, null, null, 15, 7]), - vec![vec![3], vec![20, 9], vec![15, 7]] - ); - } -} diff --git a/src/solution/s0104_maximum_depth_of_binary_tree.rs b/src/solution/s0104_maximum_depth_of_binary_tree.rs deleted file mode 100644 index 2b2a51d4..00000000 --- a/src/solution/s0104_maximum_depth_of_binary_tree.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [104] Maximum Depth of Binary Tree - * - * Given a binary tree, find its maximum depth. - * - * The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * Given binary tree [3,9,20,null,null,15,7], - * - * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * return its depth = 3. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/maximum-depth-of-binary-tree/ -// discuss: https://leetcode.com/problems/maximum-depth-of-binary-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn max_depth(root: Option>>) -> i32 { - let mut max = 0; - Solution::depth_helper(root.as_ref(), &mut max, 0); - max - } - - fn depth_helper(root: Option<&Rc>>, max: &mut i32, curr: i32) { - if let Some(node) = root { - *max = i32::max(*max, curr + 1); - Solution::depth_helper(node.borrow().left.as_ref(), max, curr + 1); - Solution::depth_helper(node.borrow().right.as_ref(), max, curr + 1); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_104() { - assert_eq!(Solution::max_depth(tree![]), 0); - assert_eq!(Solution::max_depth(tree![3, 9, 20, null, null, 15, 7]), 3); - } -} diff --git a/src/solution/s0105_construct_binary_tree_from_preorder_and_inorder_traversal.rs b/src/solution/s0105_construct_binary_tree_from_preorder_and_inorder_traversal.rs deleted file mode 100644 index 809ae654..00000000 --- a/src/solution/s0105_construct_binary_tree_from_preorder_and_inorder_traversal.rs +++ /dev/null @@ -1,71 +0,0 @@ -/** - * [105] Construct Binary Tree from Preorder and Inorder Traversal - * - * Given preorder and inorder traversal of a tree, construct the binary tree. - * - * Note:
- * You may assume that duplicates do not exist in the tree. - * - * For example, given - * - * - * preorder = [3,9,20,15,7] - * inorder = [9,3,15,20,7] - * - * Return the following binary tree: - * - * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ -// discuss: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn build_tree(preorder: Vec, inorder: Vec) -> Option>> { - Solution::build_tree_helper(&preorder[..], &inorder[..]) - } - - fn build_tree_helper(preorder: &[i32], inorder: &[i32]) -> Option>> { - if preorder.is_empty() { - return None; - } - let root_idx = inorder.iter().position(|&v| v == preorder[0]).unwrap(); - Some(Rc::new(RefCell::new(TreeNode { - val: preorder[0], - left: Solution::build_tree_helper(&preorder[1..root_idx + 1], &inorder[0..root_idx]), - right: Solution::build_tree_helper(&preorder[root_idx + 1..], &inorder[root_idx + 1..]), - }))) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_105() { - assert_eq!( - Solution::build_tree(vec![3, 9, 20, 15, 7], vec![9, 3, 15, 20, 7]), - tree![3, 9, 20, null, null, 15, 7] - ); - assert_eq!( - Solution::build_tree(vec![3, 20, 7], vec![3, 20, 7]), - tree![3, null, 20, null, 7] - ); - assert_eq!(Solution::build_tree(vec![], vec![]), tree![]); - } -} diff --git a/src/solution/s0106_construct_binary_tree_from_inorder_and_postorder_traversal.rs b/src/solution/s0106_construct_binary_tree_from_inorder_and_postorder_traversal.rs deleted file mode 100644 index df215bbc..00000000 --- a/src/solution/s0106_construct_binary_tree_from_inorder_and_postorder_traversal.rs +++ /dev/null @@ -1,78 +0,0 @@ -/** - * [106] Construct Binary Tree from Inorder and Postorder Traversal - * - * Given inorder and postorder traversal of a tree, construct the binary tree. - * - * Note:
- * You may assume that duplicates do not exist in the tree. - * - * For example, given - * - * - * inorder = [9,3,15,20,7] - * postorder = [9,15,7,20,3] - * - * Return the following binary tree: - * - * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ -// discuss: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn build_tree(inorder: Vec, postorder: Vec) -> Option>> { - Solution::build_tree_helper(&postorder[..], &inorder[..]) - } - - fn build_tree_helper(postorder: &[i32], inorder: &[i32]) -> Option>> { - if postorder.is_empty() { - return None; - } - let root_idx = inorder - .iter() - .position(|v| v == postorder.last().unwrap()) - .unwrap(); - Some(Rc::new(RefCell::new(TreeNode { - val: *postorder.last().unwrap(), - left: Solution::build_tree_helper(&postorder[0..root_idx], &inorder[0..root_idx]), - right: Solution::build_tree_helper( - &postorder[root_idx..postorder.len() - 1], - &inorder[root_idx + 1..], - ), - }))) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_106() { - assert_eq!( - Solution::build_tree(vec![9, 3, 15, 20, 7], vec![9, 15, 7, 20, 3]), - tree![3, 9, 20, null, null, 15, 7] - ); - assert_eq!( - Solution::build_tree(vec![3, 20, 7], vec![7, 20, 3]), - tree![3, null, 20, null, 7] - ); - assert_eq!(Solution::build_tree(vec![], vec![]), tree![]); - } -} diff --git a/src/solution/s0107_binary_tree_level_order_traversal_ii.rs b/src/solution/s0107_binary_tree_level_order_traversal_ii.rs deleted file mode 100644 index 4b74e0d7..00000000 --- a/src/solution/s0107_binary_tree_level_order_traversal_ii.rs +++ /dev/null @@ -1,82 +0,0 @@ -/** - * [107] Binary Tree Level Order Traversal II - * - * Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). - * - * - * For example:
- * Given binary tree [3,9,20,null,null,15,7],
- * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * - * - * return its bottom-up level order traversal as:
- * - * [ - * [15,7], - * [9,20], - * [3] - * ] - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ -// discuss: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn level_order_bottom(root: Option>>) -> Vec> { - let mut res = Vec::new(); - let mut current_level = 0; - if root.is_none() { - return res; - } - let mut deq = VecDeque::new(); - deq.push_back((0, root.clone())); - let mut vec = Vec::new(); - while !deq.is_empty() { - if let Some((level, Some(node))) = deq.pop_front() { - deq.push_back((level + 1, node.borrow().left.clone())); - deq.push_back((level + 1, node.borrow().right.clone())); - if level > current_level { - res.push(vec); - vec = Vec::new(); - current_level = level; - } - vec.push(node.borrow().val); - } - } - if !vec.is_empty() { - res.push(vec) - } - res.reverse(); - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_107() { - assert_eq!( - Solution::level_order_bottom(tree![3, 9, 20, null, null, 15, 7]), - vec![vec![15, 7], vec![9, 20], vec![3],] - ); - } -} diff --git a/src/solution/s0108_convert_sorted_array_to_binary_search_tree.rs b/src/solution/s0108_convert_sorted_array_to_binary_search_tree.rs deleted file mode 100644 index 2cefab5c..00000000 --- a/src/solution/s0108_convert_sorted_array_to_binary_search_tree.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [108] Convert Sorted Array to Binary Search Tree - * - * Given an array where elements are sorted in ascending order, convert it to a height balanced BST. - * - * For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. - * - * Example: - * - * - * Given the sorted array: [-10,-3,0,5,9], - * - * One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST: - * - * 0 - * / \ - * -3 9 - * / / - * -10 5 - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ -// discuss: https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn sorted_array_to_bst(nums: Vec) -> Option>> { - Solution::bst_helper(&nums[..]) - } - - fn bst_helper(nums: &[i32]) -> Option>> { - if nums.is_empty() { - return None; - } - Some(Rc::new(RefCell::new(TreeNode { - val: nums[nums.len() / 2], - left: Solution::bst_helper(&nums[0..(nums.len() / 2)]), - right: Solution::bst_helper(&nums[(nums.len() / 2 + 1)..]), - }))) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_108() { - assert_eq!( - Solution::sorted_array_to_bst(vec![-10, -3, 0, 5, 9]), - tree![0, -3, 9, -10, null, 5] - ); - assert_eq!(Solution::sorted_array_to_bst(vec![]), tree![]); - } -} diff --git a/src/solution/s0109_convert_sorted_list_to_binary_search_tree.rs b/src/solution/s0109_convert_sorted_list_to_binary_search_tree.rs deleted file mode 100644 index e2070965..00000000 --- a/src/solution/s0109_convert_sorted_list_to_binary_search_tree.rs +++ /dev/null @@ -1,70 +0,0 @@ -/** - * [109] Convert Sorted List to Binary Search Tree - * - * Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. - * - * For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. - * - * Example: - * - * - * Given the sorted linked list: [-10,-3,0,5,9], - * - * One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST: - * - * 0 - * / \ - * -3 9 - * / / - * -10 5 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ -// discuss: https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn sorted_list_to_bst(head: Option>) -> Option>> { - let mut arr = Vec::new(); - let mut head = head; - while let Some(node) = head { - arr.push(node.val); - head = node.next; - } - Solution::bst_helper(&arr[..]) - } - - fn bst_helper(nums: &[i32]) -> Option>> { - if nums.is_empty() { - return None; - } - Some(Rc::new(RefCell::new(TreeNode { - val: nums[nums.len() / 2], - left: Solution::bst_helper(&nums[0..(nums.len() / 2)]), - right: Solution::bst_helper(&nums[(nums.len() / 2 + 1)..]), - }))) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_109() { - assert_eq!( - Solution::sorted_list_to_bst(linked![-10, -3, 0, 5, 9]), - tree![0, -3, 9, -10, null, 5] - ); - } -} diff --git a/src/solution/s0110_balanced_binary_tree.rs b/src/solution/s0110_balanced_binary_tree.rs deleted file mode 100644 index c3171126..00000000 --- a/src/solution/s0110_balanced_binary_tree.rs +++ /dev/null @@ -1,97 +0,0 @@ -/** - * [110] Balanced Binary Tree - * - * Given a binary tree, determine if it is height-balanced. - * - * For this problem, a height-balanced binary tree is defined as: - * - *
- * a binary tree in which the depth of the two subtrees of every node never differ by more than 1. - *
- * - * Example 1: - * - * Given the following tree [3,9,20,null,null,15,7]: - * - * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * Return true.
- *
- * Example 2: - * - * Given the following tree [1,2,2,3,3,null,null,4,4]: - * - * - * 1 - * / \ - * 2 2 - * / \ - * 3 3 - * / \ - * 4 4 - * - * - * Return false. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/balanced-binary-tree/ -// discuss: https://leetcode.com/problems/balanced-binary-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn is_balanced(root: Option>>) -> bool { - Solution::balanced_helper(root.as_ref()).is_some() - } - - fn balanced_helper(root: Option<&Rc>>) -> Option { - if let Some(node) = root { - let pair = ( - Solution::balanced_helper(node.borrow().left.as_ref()), - Solution::balanced_helper(node.borrow().right.as_ref()), - ); - match pair { - (Some(left), Some(right)) => { - if i32::abs(left - right) < 2 { - return Some(i32::max(left, right) + 1); - } else { - return None; - } - } - _ => return None, - } - } else { - Some(0) - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_110() { - assert_eq!(Solution::is_balanced(tree![]), true); - assert_eq!( - Solution::is_balanced(tree![3, 9, 20, null, null, 15, 7]), - true - ); - assert_eq!( - Solution::is_balanced(tree![1, 2, 2, 3, 3, null, null, 4, 4]), - false - ); - } -} diff --git a/src/solution/s0111_minimum_depth_of_binary_tree.rs b/src/solution/s0111_minimum_depth_of_binary_tree.rs deleted file mode 100644 index a73e49f0..00000000 --- a/src/solution/s0111_minimum_depth_of_binary_tree.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [111] Minimum Depth of Binary Tree - * - * Given a binary tree, find its minimum depth. - * - * The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * Given binary tree [3,9,20,null,null,15,7], - * - * - * 3 - * / \ - * 9 20 - * / \ - * 15 7 - * - * return its minimum depth = 2. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/minimum-depth-of-binary-tree/ -// discuss: https://leetcode.com/problems/minimum-depth-of-binary-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn min_depth(root: Option>>) -> i32 { - if root.is_none() { - return 0; - } - let mut deq = VecDeque::new(); - deq.push_back((1, root.clone())); - while !deq.is_empty() { - if let Some((level, Some(node))) = deq.pop_front() { - if node.borrow().left.is_none() && node.borrow().right.is_none() { - return level; - } - deq.push_back((level + 1, node.borrow().left.clone())); - deq.push_back((level + 1, node.borrow().right.clone())); - } - } - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_111() { - assert_eq!(Solution::min_depth(tree![3, 9, 20, null, null, 15, 7]), 2); - } -} diff --git a/src/solution/s0112_path_sum.rs b/src/solution/s0112_path_sum.rs deleted file mode 100644 index e2e0a235..00000000 --- a/src/solution/s0112_path_sum.rs +++ /dev/null @@ -1,80 +0,0 @@ -/** - * [112] Path Sum - * - * Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * Given the below binary tree and sum = 22, - * - * - * 5 - * / \ - * 4 8 - * / / \ - * 11 13 4 - * / \ \ - * 7 2 1 - * - * - * return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/path-sum/ -// discuss: https://leetcode.com/problems/path-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn has_path_sum(root: Option>>, sum: i32) -> bool { - if root.is_none() { - return false; - } - let mut deq = VecDeque::new(); - deq.push_back((0, root.unwrap().clone())); - while !deq.is_empty() { - if let Some((acc, node)) = deq.pop_front() { - let acc = acc + node.borrow().val; - if node.borrow().left.is_none() && node.borrow().right.is_none() { - if acc == sum { - return true; - } - } else { - if node.borrow().left.is_some() { - deq.push_back((acc, node.borrow().left.as_ref().unwrap().clone())) - }; - if node.borrow().right.is_some() { - deq.push_back((acc, node.borrow().right.as_ref().unwrap().clone())) - }; - } - } - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_112() { - assert_eq!( - Solution::has_path_sum( - tree![5, 4, 8, 11, null, 13, 4, 7, 2, null, null, null, 1], - 22 - ), - true - ); - } -} diff --git a/src/solution/s0113_path_sum_ii.rs b/src/solution/s0113_path_sum_ii.rs deleted file mode 100644 index 106d1921..00000000 --- a/src/solution/s0113_path_sum_ii.rs +++ /dev/null @@ -1,86 +0,0 @@ -/** - * [113] Path Sum II - * - * Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * Given the below binary tree and sum = 22, - * - * - * 5 - * / \ - * 4 8 - * / / \ - * 11 13 4 - * / \ / \ - * 7 2 5 1 - * - * - * Return: - * - * - * [ - * [5,4,11,2], - * [5,8,4,5] - * ] - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/path-sum-ii/ -// discuss: https://leetcode.com/problems/path-sum-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn path_sum(root: Option>>, sum: i32) -> Vec> { - let mut res = Vec::new(); - if root.is_none() { - return res; - } - let mut deq = VecDeque::new(); - deq.push_back((0, Vec::new(), root.clone())); - while !deq.is_empty() { - if let Some((acc, mut vec, Some(node))) = deq.pop_front() { - let acc = acc + node.borrow().val; - vec.push(node.borrow().val); - if node.borrow().left.is_none() && node.borrow().right.is_none() { - if acc == sum { - res.push(vec); - } - } else { - if node.borrow().left.is_some() { - deq.push_back((acc, vec.clone(), node.borrow().left.clone())); - } - if node.borrow().right.is_some() { - deq.push_back((acc, vec.clone(), node.borrow().right.clone())); - } - } - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_113() { - assert_eq!( - Solution::path_sum(tree![5, 4, 8, 11, null, 13, 4, 7, 2, null, null, 5, 1], 22), - vec![vec![5, 4, 11, 2], vec![5, 8, 4, 5]] - ) - } -} diff --git a/src/solution/s0114_flatten_binary_tree_to_linked_list.rs b/src/solution/s0114_flatten_binary_tree_to_linked_list.rs deleted file mode 100644 index 63668c47..00000000 --- a/src/solution/s0114_flatten_binary_tree_to_linked_list.rs +++ /dev/null @@ -1,90 +0,0 @@ -/** - * [114] Flatten Binary Tree to Linked List - * - * Given a binary tree, flatten it to a linked list in-place. - * - * For example, given the following tree: - * - * - * 1 - * / \ - * 2 5 - * / \ \ - * 3 4 6 - * - * - * The flattened tree should look like: - * - * - * 1 - * \ - * 2 - * \ - * 3 - * \ - * 4 - * \ - * 5 - * \ - * 6 - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/ -// discuss: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn flatten(root: &mut Option>>) { - Solution::flatten_helper(root.clone()); - } - - fn flatten_helper(root: Option>>) -> Option>> { - if let Some(node) = root { - let tail = Solution::flatten_helper(node.borrow().left.clone()); - Solution::flatten_helper(node.borrow_mut().right.clone()); - let mut right = node.borrow().right.clone(); - let mut ptr = node.clone(); - if let Some(tail) = tail { - let head = node.borrow_mut().left.take(); - node.borrow_mut().right = head; - tail.borrow_mut().right = right.clone(); - ptr = tail.clone(); - } - while let Some(next) = ptr.clone().borrow().right.clone() { - ptr = next - } - Some(ptr) - } else { - None - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_114() { - let mut tree = tree![1, 2, 5, 3, 4, null, 6]; - Solution::flatten(&mut tree); - assert_eq!(tree, tree![1, null, 2, null, 3, null, 4, null, 5, null, 6]); - - let mut tree = tree![1, 2, null, 3]; - Solution::flatten(&mut tree); - assert_eq!(tree, tree![1, null, 2, null, 3]); - - let mut tree = tree![1, null, 2, 3]; - Solution::flatten(&mut tree); - assert_eq!(tree, tree![1, null, 2, null, 3]); - } -} diff --git a/src/solution/s0115_distinct_subsequences.rs b/src/solution/s0115_distinct_subsequences.rs deleted file mode 100644 index 7af3d08b..00000000 --- a/src/solution/s0115_distinct_subsequences.rs +++ /dev/null @@ -1,122 +0,0 @@ -/** - * [115] Distinct Subsequences - * - * Given a string S and a string T, count the number of distinct subsequences of S which equals T. - * - * A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). - * - * Example 1: - * - * - * Input: S = "rabbbit", T = "rabbit" - * Output: 3 - * Explanation: - * - * As shown below, there are 3 ways you can generate "rabbit" from S. - * (The caret symbol ^ means the chosen letters) - * - * rabbbit - * ^^^^ ^^ - * rabbbit - * ^^ ^^^^ - * rabbbit - * ^^^ ^^^ - * - * - * Example 2: - * - * - * Input: S = "babgbag", T = "bag" - * Output: 5 - * Explanation: - * - * As shown below, there are 5 ways you can generate "bag" from S. - * (The caret symbol ^ means the chosen letters) - * - * babgbag - * ^^ ^ - * babgbag - * ^^ ^ - * babgbag - * ^ ^^ - * babgbag - * ^ ^^ - * babgbag - * ^^^ - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/distinct-subsequences/ -// discuss: https://leetcode.com/problems/distinct-subsequences/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -首先想到 DFS. 但这里 DFS 有重复计算, 因为我们不需要列出所有的路径, 复杂度可以考虑 "aaaaaaaaaaaaaaaaaaaa" -里找 "aaaaaaaaa", 直接搜索的话复杂度是指数级的(阶乘), 原因很明显, 这本身是个排列组合, 可以套 combination 公式 -20! / ((20-10)! * 10!) 得到结果是 184756 - -要把复杂度从指数级降下来, 那么必须干掉重复计算, 那就想到 memorize, 想到 memorize 就想到 DP 和 Bottom-Up 递推, -回顾一下 #62 和 #63 这两个问题 (unique paths), 使用的是Bottom-Up DP, 到达每个格子的可能路径是上下两个格子的 -可能路径的和. 这里就跳过了很多的计算, 不需要把每条路径都遍历出来了. 在 unique paths 问题中, 到达右下角的路径数的 -子问题是到达右下角左侧格子的路径数以及到达右下角上侧格子的路径数. 这个问题也是一样的道理, s 中找子序列 t: - -* s[0..i] 包含的 t 序列数就是所有 s[0..j] (j < i) 包含的 t[0..t.len()-1] 的序列数 - -以 babgbag 中找 bag 为例, 做一次 Bottom-Up 递推: - - b a b g b a g -b 1 1 1 3 找 'b' 这个子序列, 那么以 [0, 2, 4] 这三个下标结尾各有一种 -a 1 3 4 找 'ba' 这个子序列, 那么以 1 结尾有1种(0 < 1), 以 5 结尾有 3 种 (0,2,4 都 < 5) -g 1 4 5 同理, 以 3 结尾有 1 种, 以 6 结尾有 4 种, 共 5 种 - -显然, 计算第 N 行时只依赖第 N-1 行的结果, 因此我们不需要存储整个矩阵, 只需要存储一行即可 - -时间复杂度是 O(M*N), 空间复杂度是 O(M) -*/ -impl Solution { - pub fn num_distinct(s: String, t: String) -> i32 { - let s = s.chars().collect::>(); - let mut cache = vec![0; s.len()]; - for (i, ch) in t.chars().into_iter().enumerate() { - let mut acc = 0; - // first char initialization - if i == 0 { - for i in 0..s.len() { - if ch == s[i] { - cache[i] = 1 - } - } - continue; - } - for i in 0..s.len() { - let new_acc = acc + cache[i]; - cache[i] = if s[i] == ch { acc } else { 0 }; - acc = new_acc; - } - } - cache.into_iter().fold(0, |acc, x| acc + x) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_115() { - //assert_eq!(Solution::num_distinct("rabbbit".to_owned(), "rabbit".to_owned()), 3); - assert_eq!( - Solution::num_distinct("babgbag".to_owned(), "bag".to_owned()), - 5 - ); - assert_eq!( - Solution::num_distinct("aaaaaaaaaaaaaaaaaaaa".to_owned(), "aaaaaaaaaa".to_owned()), - 184756 - ); - } -} diff --git a/src/solution/s0118_pascals_triangle.rs b/src/solution/s0118_pascals_triangle.rs deleted file mode 100644 index d593242d..00000000 --- a/src/solution/s0118_pascals_triangle.rs +++ /dev/null @@ -1,70 +0,0 @@ -/** - * [118] Pascal's Triangle - * - * Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. - * - *
- * In Pascal's triangle, each number is the sum of the two numbers directly above it. - * - * Example: - * - * - * Input: 5 - * Output: - * [ - * [1], - * [1,1], - * [1,2,1], - * [1,3,3,1], - * [1,4,6,4,1] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/pascals-triangle/ -// discuss: https://leetcode.com/problems/pascals-triangle/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn generate(num_rows: i32) -> Vec> { - let mut res = Vec::new(); - if num_rows < 1 { - return res; - } - let mut curr = vec![1]; - for _ in 0..num_rows { - let mut next = vec![1; curr.len() + 1]; - for i in 1..curr.len() { - next[i] = curr[i - 1] + curr[i]; - } - res.push(curr); - curr = next; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_118() { - assert_eq!(Solution::generate(1), vec![vec![1]]); - assert_eq!( - Solution::generate(5), - vec![ - vec![1], - vec![1, 1], - vec![1, 2, 1], - vec![1, 3, 3, 1], - vec![1, 4, 6, 4, 1] - ] - ); - } -} diff --git a/src/solution/s0119_pascals_triangle_ii.rs b/src/solution/s0119_pascals_triangle_ii.rs deleted file mode 100644 index 588a9835..00000000 --- a/src/solution/s0119_pascals_triangle_ii.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [119] Pascal's Triangle II - * - * Given a non-negative index k where k ≤ 33, return the k^th index row of the Pascal's triangle. - * - * Note that the row index starts from 0. - * - *
- * In Pascal's triangle, each number is the sum of the two numbers directly above it. - * - * Example: - * - * - * Input: 3 - * Output: [1,3,3,1] - * - * - * Follow up: - * - * Could you optimize your algorithm to use only O(k) extra space? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/pascals-triangle-ii/ -// discuss: https://leetcode.com/problems/pascals-triangle-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -in-place algorithm - -1 1 1 1 1 -1 2 1 1 1 -1 3 3 1 1 -1 4 6 4 1 -*/ -impl Solution { - pub fn get_row(row_index: i32) -> Vec { - let mut curr = vec![1; (row_index + 1) as usize]; - for i in 0..row_index + 1 { - let mut prev = 1; - for j in 1..i { - let temp = curr[j as usize]; - curr[j as usize] = temp + prev; - prev = temp; - } - } - curr - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_119() { - assert_eq!(Solution::get_row(0), vec![1]); - assert_eq!(Solution::get_row(4), vec![1, 4, 6, 4, 1]) - } -} diff --git a/src/solution/s0120_triangle.rs b/src/solution/s0120_triangle.rs deleted file mode 100644 index 93c38d33..00000000 --- a/src/solution/s0120_triangle.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [120] Triangle - * - * Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. - * - * For example, given the following triangle - * - * - * [ - * [2], - * [3,4], - * [6,5,7], - * [4,1,8,3] - * ] - * - * - * The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). - * - * Note: - * - * Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/triangle/ -// discuss: https://leetcode.com/problems/triangle/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn minimum_total(triangle: Vec>) -> i32 { - let mut cache = vec![0; triangle.len()]; - for row in triangle.iter() { - let mut prev = 0; - for i in 0..row.len() { - let temp = cache[i]; - cache[i] = if i == 0 { - cache[i] - } else if i == row.len() - 1 { - prev - } else { - i32::min(cache[i], prev) - } + row[i]; - prev = temp; - } - } - cache - .into_iter() - .fold(i32::max_value(), |min, x| i32::min(min, x)) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_120() { - assert_eq!( - Solution::minimum_total(vec![vec![2], vec![3, 4], vec![6, 5, 7], vec![4, 1, 8, 3]]), - 11 - ) - } -} diff --git a/src/solution/s0121_best_time_to_buy_and_sell_stock.rs b/src/solution/s0121_best_time_to_buy_and_sell_stock.rs deleted file mode 100644 index d5d3746d..00000000 --- a/src/solution/s0121_best_time_to_buy_and_sell_stock.rs +++ /dev/null @@ -1,62 +0,0 @@ -/** - * [121] Best Time to Buy and Sell Stock - * - * Say you have an array for which the i^th element is the price of a given stock on day i. - * - * If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit. - * - * Note that you cannot sell a stock before you buy one. - * - * Example 1: - * - * - * Input: [7,1,5,3,6,4] - * Output: 5 - * Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. - * Not 7-1 = 6, as selling price needs to be larger than buying price. - * - * - * Example 2: - * - * - * Input: [7,6,4,3,1] - * Output: 0 - * Explanation: In this case, no transaction is done, i.e. max profit = 0. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ -// discuss: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn max_profit(prices: Vec) -> i32 { - let mut max = 0; - let mut curr = 0; - for i in 1..prices.len() { - curr = curr + prices[i] - prices[i - 1]; - if curr <= 0 { - curr = 0; - } else { - max = i32::max(max, curr); - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_121() { - assert_eq!(Solution::max_profit(vec![7, 1, 5, 3, 6, 4]), 5); - assert_eq!(Solution::max_profit(vec![7, 6, 4, 3, 1]), 0); - } -} diff --git a/src/solution/s0122_best_time_to_buy_and_sell_stock_ii.rs b/src/solution/s0122_best_time_to_buy_and_sell_stock_ii.rs deleted file mode 100644 index 803c71b6..00000000 --- a/src/solution/s0122_best_time_to_buy_and_sell_stock_ii.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [122] Best Time to Buy and Sell Stock II - * - * Say you have an array for which the i^th element is the price of a given stock on day i. - * - * Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times). - * - * Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again). - * - * Example 1: - * - * - * Input: [7,1,5,3,6,4] - * Output: 7 - * Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. - * Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. - * - * - * Example 2: - * - * - * Input: [1,2,3,4,5] - * Output: 4 - * Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. - * Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are - * engaging multiple transactions at the same time. You must sell before buying again. - * - * - * Example 3: - * - * - * Input: [7,6,4,3,1] - * Output: 0 - * Explanation: In this case, no transaction is done, i.e. max profit = 0. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ -// discuss: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn max_profit(prices: Vec) -> i32 { - let mut max = 0; - for i in 1..prices.len() { - if prices[i] > prices[i - 1] { - max += prices[i] - prices[i - 1]; - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_122() { - assert_eq!(Solution::max_profit(vec![7, 1, 5, 3, 6, 4]), 7); - assert_eq!(Solution::max_profit(vec![1, 2, 3, 4, 5]), 4); - } -} diff --git a/src/solution/s0123_best_time_to_buy_and_sell_stock_iii.rs b/src/solution/s0123_best_time_to_buy_and_sell_stock_iii.rs deleted file mode 100644 index bb068273..00000000 --- a/src/solution/s0123_best_time_to_buy_and_sell_stock_iii.rs +++ /dev/null @@ -1,106 +0,0 @@ -/** - * [123] Best Time to Buy and Sell Stock III - * - * Say you have an array for which the i^th element is the price of a given stock on day i. - * - * Design an algorithm to find the maximum profit. You may complete at most two transactions. - * - * Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again). - * - * Example 1: - * - * - * Input: [3,3,5,0,0,3,1,4] - * Output: 6 - * Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. - * Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3. - * - * Example 2: - * - * - * Input: [1,2,3,4,5] - * Output: 4 - * Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. - * Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are - * engaging multiple transactions at the same time. You must sell before buying again. - * - * - * Example 3: - * - * - * Input: [7,6,4,3,1] - * Output: 0 - * Explanation: In this case, no transaction is done, i.e. max profit = 0. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/ -// discuss: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -先考虑只进行 1 次交易的情况, 我们求以 i *为售出点*, 只进行 1 次交易获得的最大利润, 那么: - -f[i] = if f[i-1] > 0 { f[i-1] } else { 0 } + prices[i] - prices[i-1] - -这很容易解, 解完之后找出 f 里的最大值即可, 但这不容易推广到 K 次交易的情况, 因为这时 f[i] 不代表到 i *为止*的最大利润, 无法作为单独的交易帮助递推 -(到 i 为止的含义是售出点可以在 [0,i] 之间) - -我们可以稍作改进, 变成求以 i 为结束点, 只进行 1 次交易获得的最大利润, 那么: - -f[i] = max( - f[i-1], - prices[i] - min(prices[j] { j in [0, i-1] }) -) - -这仍然是一个 O(N) 的解法, 因为 min(prices[j] { j in [0, i-1] }) 不需要遍历, 可以在递推过程中直接维护好 - -现在再推广到进行 K 次交易的情况, 那我们要求以 i 为结束点, 进行 k 次交易获得的最大利润, 这时有了变化, 我们可以在 j 之前再进行 K - 1 次交易: - -f[k, i] = max( - f[k, i-1], - prices[i] + max(f[k-1, j] - prices[j]) { j in [0, i-1] } ) -) - -显然, f[0, i] = 0, f[k, 0] = 0 - -这个算法可以形象地描述一下, 在 k = 1 时, 我们每次要找的就是 i 之前的最低谷点作为这次交易的开始点 j, 而当 k > 1 时, -我们 i 之前就有可能已经进行过交易了, 这时我们在找开始点 j 时, 就要同时考虑 "直到 j 为止, k-1 次交易的最大收益" - "j 本身的值". 以此来找到一个最佳点 j - -在实现时, 假如用 Bottom-Up 递推, 那么只需要维护一个 vec[i], 因为每轮递推时只会考虑上一轮的数据, 我们可以复用这个 O(N) 的额外存储空间 -*/ -impl Solution { - pub fn max_profit(prices: Vec) -> i32 { - if prices.is_empty() { - return 0; - } - let max_trans = 2; - let mut cache = vec![0; prices.len()]; - for trans in 0..max_trans { - // best_by_in 维护了考虑前 N 次交易的最佳的买入点, 即 max(f[k-1, j] - prices[j]) { j in [0, i-1] } - let mut best_buy_in = cache[0] - prices[0]; - for i in 1..prices.len() { - // 复用 vec 前暂存一下前一次的计算结果 - let temp = cache[i]; - cache[i] = i32::max(cache[i - 1], best_buy_in + prices[i]); - // 更新 best_buy_in - best_buy_in = i32::max(best_buy_in, temp - prices[i]); - } - } - return *cache.last().unwrap(); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_123() { - assert_eq!(Solution::max_profit(vec![3, 3, 5, 0, 0, 3, 1, 4]), 6); - } -} diff --git a/src/solution/s0124_binary_tree_maximum_path_sum.rs b/src/solution/s0124_binary_tree_maximum_path_sum.rs deleted file mode 100644 index e7c57b49..00000000 --- a/src/solution/s0124_binary_tree_maximum_path_sum.rs +++ /dev/null @@ -1,100 +0,0 @@ -/** - * [124] Binary Tree Maximum Path Sum - * - * Given a non-empty binary tree, find the maximum path sum. - * - * For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root. - * - * Example 1: - * - * - * Input: [1,2,3] - * - * 1 - * / \ - * 2 3 - * - * Output: 6 - * - * - * Example 2: - * - * - * Input: [-10,9,20,null,null,15,7] - * - * -10 - * / \ - * 9 20 - * / \ - * 15 7 - * - * Output: 42 - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-maximum-path-sum/ -// discuss: https://leetcode.com/problems/binary-tree-maximum-path-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -典型的动态规划, 我们求以 node_i 为 root 的最大和, 可以下推到求 root 的左右子树, 这里要注意, 路径是不能分叉的, 因此 -我们记 f[node] 为以 node 为根的最大和, 记 g[node] 为 node 为根, *最多连通一侧子树*的最大和 - -我们在递推时要用 g[node], f[node] 在递推过程中每次计算一下用于更新 max 即可 - -g[node_i] = node_i.val + max(g[node_i.left], g[node_i.right], 0) -f[node_i] = node_i.val + max(g[node_i.left], 0) + max(g[node_i.right], 0) - -显然, g[None] = 0 (None 即空子树), 最终计算到 g[root] 中止, f 的最大值会在计算过程中出现(注意 f[root] 不一定是最大值) - -每个 node 最大和只依赖与其左右子树的最大和, 因此 Top-down 需要 O(N) 的空间 -Bottom-up 只需要 O(1) 空间 (做后序遍历从叶节点向上递推即可) -*/ -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn max_path_sum(root: Option>>) -> i32 { - let mut max = i32::min_value(); - Solution::postorder(root.as_ref(), &mut max); - max - } - - fn postorder(root: Option<&Rc>>, max: &mut i32) -> i32 { - if let Some(node) = root { - let left = Solution::postorder(node.borrow().left.as_ref(), max); - let right = Solution::postorder(node.borrow().right.as_ref(), max); - *max = i32::max( - node.borrow().val + i32::max(left, 0) + i32::max(right, 0), - *max, - ); - node.borrow().val + i32::max(i32::max(left, right), 0) - } else { - 0 - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_124() { - assert_eq!(Solution::max_path_sum(tree![1, 2, 3]), 6); - assert_eq!( - Solution::max_path_sum(tree![-10, 9, 20, null, null, 15, 7]), - 42 - ); - assert_eq!( - Solution::max_path_sum(tree![5, 4, 8, 11, null, 13, 4, 7, 2, null, null, null, 1]), - 48 - ); - assert_eq!(Solution::max_path_sum(tree![-3]), -3); - } -} diff --git a/src/solution/s0125_valid_palindrome.rs b/src/solution/s0125_valid_palindrome.rs deleted file mode 100644 index 2f090b72..00000000 --- a/src/solution/s0125_valid_palindrome.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [125] Valid Palindrome - * - * Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. - * - * Note: For the purpose of this problem, we define empty string as valid palindrome. - * - * Example 1: - * - * - * Input: "A man, a plan, a canal: Panama" - * Output: true - * - * - * Example 2: - * - * - * Input: "race a car" - * Output: false - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/valid-palindrome/ -// discuss: https://leetcode.com/problems/valid-palindrome/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_palindrome(s: String) -> bool { - if s.is_empty() { - return true; - } - let seq = s.chars().collect::>(); - let (mut i, mut j) = (0_usize, seq.len() - 1); - while i < j { - while i < seq.len() && !seq[i].is_ascii_alphanumeric() { - i += 1; - } - while j > 0 && !seq[j].is_ascii_alphanumeric() { - j -= 1; - } - if i >= j { - break; - } - if seq[i].to_ascii_lowercase() != seq[j].to_ascii_lowercase() { - return false; - } - i += 1; - j -= 1; - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_125() { - assert_eq!( - Solution::is_palindrome("A man, a plan, a canal: Panama".to_owned()), - true - ); - assert_eq!(Solution::is_palindrome("race a car".to_owned()), false); - assert_eq!(Solution::is_palindrome("0P".to_owned()), false); - } -} diff --git a/src/solution/s0126_word_ladder_ii.rs b/src/solution/s0126_word_ladder_ii.rs deleted file mode 100644 index 32169865..00000000 --- a/src/solution/s0126_word_ladder_ii.rs +++ /dev/null @@ -1,258 +0,0 @@ -/** - * [126] Word Ladder II - * - * Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: - * - *
    - * Only one letter can be changed at a time - * Each transformed word must exist in the word list. Note that beginWord is not a transformed word. - *
- * - * Note: - * - * - * Return an empty list if there is no such transformation sequence. - * All words have the same length. - * All words contain only lowercase alphabetic characters. - * You may assume no duplicates in the word list. - * You may assume beginWord and endWord are non-empty and are not the same. - * - * - * Example 1: - * - * - * Input: - * beginWord = "hit", - * endWord = "cog", - * wordList = ["hot","dot","dog","lot","log","cog"] - * - * Output: - * [ - * ["hit","hot","dot","dog","cog"], - * ["hit","hot","lot","log","cog"] - * ] - * - * - * Example 2: - * - * - * Input: - * beginWord = "hit" - * endWord = "cog" - * wordList = ["hot","dot","dog","lot","log"] - * - * Output: [] - * - * Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. - * - * - * - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-ladder-ii/ -// discuss: https://leetcode.com/problems/word-ladder-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -假如 A 经过一个字符的变换能得到 B, 则认为 A 与 B 之间有通路, 转化为一个 BFS 找无权图最短路径的问题 - -实现时, 可以先把图构造出来, 复杂度 O(L*N^2) (L 是字符串长度), 也可以每次都回到数组里去找连通点, 时间复杂度不变 - -由于要记录所有的路径, 因此我们需要把每个点的可能前置节点都记录下来, 最后用一个 DFS 或 BFS 找出所有路径 - -暂时想不到更好的办法 -*/ - -use std::collections::HashSet; -use std::collections::VecDeque; -impl Solution { - pub fn find_ladders( - begin_word: String, - end_word: String, - word_list: Vec, - ) -> Vec> { - let mut res = Vec::new(); - let len = word_list.len(); - let target = word_list.iter().position(|s| s == &end_word); - if target.is_none() { - return res; - } - let target = target.unwrap(); - let mut deq = VecDeque::new(); - deq.push_back(target); - // paths record the distance & previous index, we use 'len' to represent empty prev - let mut paths: Vec<(i32, Vec)> = vec![(i32::max_value(), vec![]); len]; - paths[target].0 = 0; - let mut find_shortest = false; - let mut shortest = i32::max_value(); - let mut in_queue = HashSet::new(); - while let Some(i) = deq.pop_front() { - if Solution::connect(&begin_word, &word_list[i]) { - // complete the path using dfs - if paths[i].0 > shortest { - continue; - } - Solution::dfs(i, vec![begin_word.clone()], &word_list, &paths, &mut res); - shortest = paths[i].0; - find_shortest = true; - } - // we have found the shortest path, just drain all the nodes in deq - if find_shortest { - continue; - } - for j in 0..len { - if j == i { - continue; - } - if Solution::connect(&word_list[i], &word_list[j]) { - if paths[i].0 + 1 <= paths[j].0 { - let mut prev = &mut paths[j].1; - prev.push(i); - paths[j].0 = paths[i].0 + 1; - if !in_queue.contains(&j) { - deq.push_back(j); - in_queue.insert(j); - } - } - } - } - } - res - } - - fn dfs( - curr: usize, - mut path: Vec, - words: &Vec, - paths: &Vec<(i32, Vec)>, - res: &mut Vec>, - ) { - path.push(words[curr].clone()); - if paths[curr].1.is_empty() { - res.push(path); - return; - } - for &prev in paths[curr].1.iter() { - Solution::dfs(prev, path.clone(), words, paths, res); - } - } - - #[inline(always)] - fn connect(s1: &str, s2: &str) -> bool { - if s1.len() != s2.len() { - return false; - } - let mut iter1 = s1.chars().into_iter(); - let mut iter2 = s2.chars().into_iter(); - let mut diff = 0; - while let (Some(c1), Some(c2)) = (iter1.next(), iter2.next()) { - if c1 != c2 { - diff += 1; - if diff >= 2 { - return false; - } - } - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_126() { - assert_eq!( - Solution::find_ladders( - "hit".to_owned(), - "cog".to_owned(), - vec_string!["hot", "dot", "dog", "lot", "log", "cog"] - ), - vec![ - vec_string!["hit", "hot", "dot", "dog", "cog"], - vec_string!["hit", "hot", "lot", "log", "cog"], - ] - ); - assert_eq!( - Solution::find_ladders( - "cet".to_owned(), - "ism".to_owned(), - vec_string![ - "kid", "tag", "pup", "ail", "tun", "woo", "erg", "luz", "brr", "gay", "sip", - "kay", "per", "val", "mes", "ohs", "now", "boa", "cet", "pal", "bar", "die", - "war", "hay", "eco", "pub", "lob", "rue", "fry", "lit", "rex", "jan", "cot", - "bid", "ali", "pay", "col", "gum", "ger", "row", "won", "dan", "rum", "fad", - "tut", "sag", "yip", "sui", "ark", "has", "zip", "fez", "own", "ump", "dis", - "ads", "max", "jaw", "out", "btu", "ana", "gap", "cry", "led", "abe", "box", - "ore", "pig", "fie", "toy", "fat", "cal", "lie", "noh", "sew", "ono", "tam", - "flu", "mgm", "ply", "awe", "pry", "tit", "tie", "yet", "too", "tax", "jim", - "san", "pan", "map", "ski", "ova", "wed", "non", "wac", "nut", "why", "bye", - "lye", "oct", "old", "fin", "feb", "chi", "sap", "owl", "log", "tod", "dot", - "bow", "fob", "for", "joe", "ivy", "fan", "age", "fax", "hip", "jib", "mel", - "hus", "sob", "ifs", "tab", "ara", "dab", "jag", "jar", "arm", "lot", "tom", - "sax", "tex", "yum", "pei", "wen", "wry", "ire", "irk", "far", "mew", "wit", - "doe", "gas", "rte", "ian", "pot", "ask", "wag", "hag", "amy", "nag", "ron", - "soy", "gin", "don", "tug", "fay", "vic", "boo", "nam", "ave", "buy", "sop", - "but", "orb", "fen", "paw", "his", "sub", "bob", "yea", "oft", "inn", "rod", - "yam", "pew", "web", "hod", "hun", "gyp", "wei", "wis", "rob", "gad", "pie", - "mon", "dog", "bib", "rub", "ere", "dig", "era", "cat", "fox", "bee", "mod", - "day", "apr", "vie", "nev", "jam", "pam", "new", "aye", "ani", "and", "ibm", - "yap", "can", "pyx", "tar", "kin", "fog", "hum", "pip", "cup", "dye", "lyx", - "jog", "nun", "par", "wan", "fey", "bus", "oak", "bad", "ats", "set", "qom", - "vat", "eat", "pus", "rev", "axe", "ion", "six", "ila", "lao", "mom", "mas", - "pro", "few", "opt", "poe", "art", "ash", "oar", "cap", "lop", "may", "shy", - "rid", "bat", "sum", "rim", "fee", "bmw", "sky", "maj", "hue", "thy", "ava", - "rap", "den", "fla", "auk", "cox", "ibo", "hey", "saw", "vim", "sec", "ltd", - "you", "its", "tat", "dew", "eva", "tog", "ram", "let", "see", "zit", "maw", - "nix", "ate", "gig", "rep", "owe", "ind", "hog", "eve", "sam", "zoo", "any", - "dow", "cod", "bed", "vet", "ham", "sis", "hex", "via", "fir", "nod", "mao", - "aug", "mum", "hoe", "bah", "hal", "keg", "hew", "zed", "tow", "gog", "ass", - "dem", "who", "bet", "gos", "son", "ear", "spy", "kit", "boy", "due", "sen", - "oaf", "mix", "hep", "fur", "ada", "bin", "nil", "mia", "ewe", "hit", "fix", - "sad", "rib", "eye", "hop", "haw", "wax", "mid", "tad", "ken", "wad", "rye", - "pap", "bog", "gut", "ito", "woe", "our", "ado", "sin", "mad", "ray", "hon", - "roy", "dip", "hen", "iva", "lug", "asp", "hui", "yak", "bay", "poi", "yep", - "bun", "try", "lad", "elm", "nat", "wyo", "gym", "dug", "toe", "dee", "wig", - "sly", "rip", "geo", "cog", "pas", "zen", "odd", "nan", "lay", "pod", "fit", - "hem", "joy", "bum", "rio", "yon", "dec", "leg", "put", "sue", "dim", "pet", - "yaw", "nub", "bit", "bur", "sid", "sun", "oil", "red", "doc", "moe", "caw", - "eel", "dix", "cub", "end", "gem", "off", "yew", "hug", "pop", "tub", "sgt", - "lid", "pun", "ton", "sol", "din", "yup", "jab", "pea", "bug", "gag", "mil", - "jig", "hub", "low", "did", "tin", "get", "gte", "sox", "lei", "mig", "fig", - "lon", "use", "ban", "flo", "nov", "jut", "bag", "mir", "sty", "lap", "two", - "ins", "con", "ant", "net", "tux", "ode", "stu", "mug", "cad", "nap", "gun", - "fop", "tot", "sow", "sal", "sic", "ted", "wot", "del", "imp", "cob", "way", - "ann", "tan", "mci", "job", "wet", "ism", "err", "him", "all", "pad", "hah", - "hie", "aim", "ike", "jed", "ego", "mac", "baa", "min", "com", "ill", "was", - "cab", "ago", "ina", "big", "ilk", "gal", "tap", "duh", "ola", "ran", "lab", - "top", "gob", "hot", "ora", "tia", "kip", "han", "met", "hut", "she", "sac", - "fed", "goo", "tee", "ell", "not", "act", "gil", "rut", "ala", "ape", "rig", - "cid", "god", "duo", "lin", "aid", "gel", "awl", "lag", "elf", "liz", "ref", - "aha", "fib", "oho", "tho", "her", "nor", "ace", "adz", "fun", "ned", "coo", - "win", "tao", "coy", "van", "man", "pit", "guy", "foe", "hid", "mai", "sup", - "jay", "hob", "mow", "jot", "are", "pol", "arc", "lax", "aft", "alb", "len", - "air", "pug", "pox", "vow", "got", "meg", "zoe", "amp", "ale", "bud", "gee", - "pin", "dun", "pat", "ten", "mob" - ] - ), - vec![ - vec_string![ - "cet", "get", "gee", "gte", "ate", "ats", "its", "ito", "ibo", "ibm", "ism" - ], - vec_string![ - "cet", "cat", "can", "ian", "inn", "ins", "its", "ito", "ibo", "ibm", "ism" - ], - vec_string![ - "cet", "cot", "con", "ion", "inn", "ins", "its", "ito", "ibo", "ibm", "ism" - ], - ] - ); - } -} diff --git a/src/solution/s0127_word_ladder.rs b/src/solution/s0127_word_ladder.rs deleted file mode 100644 index a67465fa..00000000 --- a/src/solution/s0127_word_ladder.rs +++ /dev/null @@ -1,132 +0,0 @@ -/** - * [127] Word Ladder - * - * Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: - * - *
    - * Only one letter can be changed at a time. - * Each transformed word must exist in the word list. Note that beginWord is not a transformed word. - *
- * - * Note: - * - * - * Return 0 if there is no such transformation sequence. - * All words have the same length. - * All words contain only lowercase alphabetic characters. - * You may assume no duplicates in the word list. - * You may assume beginWord and endWord are non-empty and are not the same. - * - * - * Example 1: - * - * - * Input: - * beginWord = "hit", - * endWord = "cog", - * wordList = ["hot","dot","dog","lot","log","cog"] - * - * Output: 5 - * - * Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", - * return its length 5. - * - * - * Example 2: - * - * - * Input: - * beginWord = "hit" - * endWord = "cog" - * wordList = ["hot","dot","dog","lot","log"] - * - * Output: 0 - * - * Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-ladder/ -// discuss: https://leetcode.com/problems/word-ladder/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashSet; -use std::collections::VecDeque; -impl Solution { - pub fn ladder_length(begin_word: String, end_word: String, word_list: Vec) -> i32 { - let len = word_list.len(); - let target = word_list.iter().position(|s| s == &end_word); - if target.is_none() { - return 0; - } - let target = target.unwrap(); - let mut deq = VecDeque::new(); - let mut distance = vec![0; len]; - let mut remain = (0..len).collect::>(); - deq.push_back(target); - remain.remove(&target); - while let Some(i) = deq.pop_front() { - if Solution::connect(&begin_word, &word_list[i]) { - return distance[i] + 2; - } - remain.retain(|&j| { - if Solution::connect(&word_list[i], &word_list[j]) { - distance[j] = distance[i] + 1; - deq.push_back(j); - false - } else { - true - } - }); - } - 0 - } - - #[inline(always)] - fn connect(s1: &str, s2: &str) -> bool { - if s1.len() != s2.len() { - return false; - } - let mut iter1 = s1.chars().into_iter(); - let mut iter2 = s2.chars().into_iter(); - let mut diff = 0; - while let (Some(c1), Some(c2)) = (iter1.next(), iter2.next()) { - if c1 != c2 { - diff += 1; - if diff >= 2 { - return false; - } - } - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_127() { - assert_eq!( - Solution::ladder_length( - "hit".to_owned(), - "cog".to_owned(), - vec_string!["hot", "dot", "dog", "lot", "log", "cog"] - ), - 5 - ); - assert_eq!( - Solution::ladder_length( - "hit".to_owned(), - "cog".to_owned(), - vec_string!["hot", "dot", "dog", "lot", "log"] - ), - 0 - ); - } -} diff --git a/src/solution/s0128_longest_consecutive_sequence.rs b/src/solution/s0128_longest_consecutive_sequence.rs deleted file mode 100644 index f0e0806f..00000000 --- a/src/solution/s0128_longest_consecutive_sequence.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [128] Longest Consecutive Sequence - * - * Given an unsorted array of integers, find the length of the longest consecutive elements sequence. - * - * Your algorithm should run in O(n) complexity. - * - * Example: - * - * - * Input: [100, 4, 200, 1, 3, 2] - * Output: 4 - * Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-consecutive-sequence/ -// discuss: https://leetcode.com/problems/longest-consecutive-sequence/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -要找到连续子串, 基本策略就是对每个 num, 判断 num+1, num+2, num+3... 是否在数组中, 直到不再连续 - -工程里写的话用排序是最清晰可维护的(需求变了很好改), 排序之后查找 num+1 是否存在就只需要 O(1) 的复杂度了: -看下一个元素是不是 num+1 即可 - -但题目一定要求 O(N) 的解法, 只能想些奇怪的办法了, HashSet 也能达到 O(1) 的查找效率. 但假如对每个元素 -都做一遍, 最差就是 O(N^2) 了, 可以发现对于一个连续序列 1,2,3,4,5,6 我们从 1 开始查就能找到这个序列, -从 2,3,4,5,6 开始查都是在做重复计算, 因此对于一个 num, 假如 num-1 存在于 HashSet 中, 我们就不需要考虑 -它了, 因为它是一次重复的计算. -*/ -use std::collections::HashSet; -impl Solution { - pub fn longest_consecutive(nums: Vec) -> i32 { - let mut max = 0; - let nums = nums.into_iter().collect::>(); - for &num in nums.iter() { - if !nums.contains(&(num - 1)) { - let mut curr = num; - let mut curr_max = 1; - while nums.contains(&(curr + 1)) { - curr += 1; - curr_max += 1; - } - max = i32::max(curr_max, max); - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_128() { - assert_eq!(Solution::longest_consecutive(vec![100, 4, 200, 1, 3, 2]), 4) - } -} diff --git a/src/solution/s0129_sum_root_to_leaf_numbers.rs b/src/solution/s0129_sum_root_to_leaf_numbers.rs deleted file mode 100644 index 1e44fb68..00000000 --- a/src/solution/s0129_sum_root_to_leaf_numbers.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [129] Sum Root to Leaf Numbers - * - * Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. - * - * An example is the root-to-leaf path 1->2->3 which represents the number 123. - * - * Find the total sum of all root-to-leaf numbers. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * - * Input: [1,2,3] - * 1 - * / \ - * 2 3 - * Output: 25 - * Explanation: - * The root-to-leaf path 1->2 represents the number 12. - * The root-to-leaf path 1->3 represents the number 13. - * Therefore, sum = 12 + 13 = 25. - * - * Example 2: - * - * - * Input: [4,9,0,5,1] - * 4 - * / \ - * 9 0 - * / \ - * 5 1 - * Output: 1026 - * Explanation: - * The root-to-leaf path 4->9->5 represents the number 495. - * The root-to-leaf path 4->9->1 represents the number 491. - * The root-to-leaf path 4->0 represents the number 40. - * Therefore, sum = 495 + 491 + 40 = 1026. - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/sum-root-to-leaf-numbers/ -// discuss: https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn sum_numbers(root: Option>>) -> i32 { - let mut sum = 0; - if root.is_none() { - return sum; - } - let mut deq = VecDeque::new(); - deq.push_back((root.clone(), 0)); - while let Some(item) = deq.pop_front() { - if let (Some(node), acc) = item { - let acc = acc * 10 + node.borrow().val; - if node.borrow().left.is_none() && node.borrow().right.is_none() { - sum += acc; - continue; - } - deq.push_back((node.borrow().left.clone(), acc)); - deq.push_back((node.borrow().right.clone(), acc)); - } - } - sum - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_129() { - assert_eq!(Solution::sum_numbers(tree![1, 2, 3]), 25); - assert_eq!(Solution::sum_numbers(tree![4, 9, 0, 5, 1]), 1026); - } -} diff --git a/src/solution/s0130_surrounded_regions.rs b/src/solution/s0130_surrounded_regions.rs deleted file mode 100644 index 411286b4..00000000 --- a/src/solution/s0130_surrounded_regions.rs +++ /dev/null @@ -1,221 +0,0 @@ -/** - * [130] Surrounded Regions - * - * Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'. - * - * A region is captured by flipping all 'O's into 'X's in that surrounded region. - * - * Example: - * - * - * X X X X - * X O O X - * X X O X - * X O X X - * - * - * After running your function, the board should be: - * - * - * X X X X - * X X X X - * X X X X - * X O X X - * - * - * Explanation: - * - * Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/surrounded-regions/ -// discuss: https://leetcode.com/problems/surrounded-regions/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -从最外层开始, 基于为 'O' 的格子做 DFS, 将与边界连接的所有 'O' 标记为一个特殊 char, 最后将没有标记到的 'O' 全部标记为 'X' -*/ -impl Solution { - pub fn solve(board: &mut Vec>) { - if board.is_empty() || board[0].is_empty() { - return; - } - let (height, width) = (board.len(), board[0].len()); - // 遍历最外层的 4 条边 - for j in 0..width { - Solution::dfs(0, j, height, width, board); - Solution::dfs(height - 1, j, height, width, board); - } - for i in 1..height - 1 { - Solution::dfs(i, 0, height, width, board); - Solution::dfs(i, width - 1, height, width, board); - } - for k in 0..height * width { - board[k / width][k % width] = if board[k / width][k % width] == '_' { - 'O' - } else { - 'X' - } - } - } - - fn dfs(i: usize, j: usize, height: usize, width: usize, board: &mut Vec>) { - if board[i][j] == 'O' { - board[i][j] = '_'; - if i > 1 { - Solution::dfs(i - 1, j, height, width, board) - } - if j > 1 { - Solution::dfs(i, j - 1, height, width, board) - } - if i + 1 < height { - Solution::dfs(i + 1, j, height, width, board) - } - if j + 1 < width { - Solution::dfs(i, j + 1, height, width, board) - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_130() { - let mut matrix = vec![ - vec!['X', 'X', 'X', 'X'], - vec!['X', 'O', 'O', 'X'], - vec!['X', 'X', 'O', 'X'], - vec!['X', 'O', 'X', 'X'], - ]; - Solution::solve(&mut matrix); - assert_eq!( - matrix, - vec![ - vec!['X', 'X', 'X', 'X'], - vec!['X', 'X', 'X', 'X'], - vec!['X', 'X', 'X', 'X'], - vec!['X', 'O', 'X', 'X'], - ] - ); - - let mut matrix = vec![ - vec!['X', 'X', 'X', 'X'], - vec!['X', 'O', 'O', 'X'], - vec!['X', 'O', 'O', 'X'], - vec!['X', 'X', 'X', 'X'], - ]; - Solution::solve(&mut matrix); - assert_eq!( - matrix, - vec![ - vec!['X', 'X', 'X', 'X'], - vec!['X', 'X', 'X', 'X'], - vec!['X', 'X', 'X', 'X'], - vec!['X', 'X', 'X', 'X'], - ] - ); - - let mut matrix = vec![ - vec!['X', 'X', 'X', 'X'], - vec!['O', 'X', 'O', 'X'], - vec!['O', 'X', 'O', 'X'], - vec!['X', 'O', 'X', 'X'], - ]; - Solution::solve(&mut matrix); - assert_eq!( - matrix, - vec![ - vec!['X', 'X', 'X', 'X'], - vec!['O', 'X', 'X', 'X'], - vec!['O', 'X', 'X', 'X'], - vec!['X', 'O', 'X', 'X'], - ] - ); - - let mut matrix = vec![ - vec!['X', 'X', 'X', 'X', 'O', 'X'], - vec!['O', 'X', 'X', 'O', 'O', 'X'], - vec!['X', 'O', 'X', 'O', 'O', 'O'], - vec!['X', 'O', 'O', 'O', 'X', 'O'], - vec!['O', 'O', 'X', 'X', 'O', 'X'], - vec!['X', 'O', 'X', 'O', 'X', 'X'], - ]; - Solution::solve(&mut matrix); - assert_eq!( - matrix, - vec![ - vec!['X', 'X', 'X', 'X', 'O', 'X'], - vec!['O', 'X', 'X', 'O', 'O', 'X'], - vec!['X', 'O', 'X', 'O', 'O', 'O'], - vec!['X', 'O', 'O', 'O', 'X', 'O'], - vec!['O', 'O', 'X', 'X', 'X', 'X'], - vec!['X', 'O', 'X', 'O', 'X', 'X'], - ] - ); - - let mut matrix = vec![ - vec![ - 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'O', 'O', 'O', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'O', 'O', 'X', 'O', 'X', 'O', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'O', 'X', 'O', 'X', 'O', 'O', 'O', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'O', 'O', 'O', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X', - ], - ]; - Solution::solve(&mut matrix); - assert_eq!( - matrix, - vec![ - vec![ - 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X' - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'O', 'O', 'O', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X' - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'O', 'O', 'X', 'O', 'X', 'O', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X' - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'O', 'X', 'O', 'X', 'O', 'O', 'O', 'X', 'X', - 'X', 'X', 'X', 'X' - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'O', 'O', 'O', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X' - ], - vec![ - 'X', 'X', 'X', 'X', 'X', 'O', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', - 'X', 'X', 'X', 'X' - ] - ] - ); - } -} diff --git a/src/solution/s0131_palindrome_partitioning.rs b/src/solution/s0131_palindrome_partitioning.rs deleted file mode 100644 index 22a92dfd..00000000 --- a/src/solution/s0131_palindrome_partitioning.rs +++ /dev/null @@ -1,120 +0,0 @@ -/** - * [131] Palindrome Partitioning - * - * Given a string s, partition s such that every substring of the partition is a palindrome. - * - * Return all possible palindrome partitioning of s. - * - * Example: - * - * - * Input: "aab" - * Output: - * [ - * ["aa","b"], - * ["a","a","b"] - * ] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/palindrome-partitioning/ -// discuss: https://leetcode.com/problems/palindrome-partitioning/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -记 n 个字符的回文拆分方式是 f(n) 种, 则: - -f(n) = (0..n).iter().fold(0, |acc, i| { - if is_palindrome(s[i..n]) { acc + f(i-1) } else { acc } -}) - -按这种方式向上递推即可, 时间复杂度为 O(N^3), 空间复杂度 O(N), 显然, is_palindrome 这一步仍然有重复计算 - -is_palindrome(s[i..n]) = s[i] == s[n] && is_palindrome(s[i+1..n-1]) - -存储所有 i, n 的 is_palindrome 结果, 则可以优化 is_palindrome 的时间到 O(1) - -最后的复杂度: 时间 O(N^2), 空间 O(N^2) -*/ -impl Solution { - pub fn partition(s: String) -> Vec> { - let s = s.chars().collect::>(); - if s.is_empty() { - return Vec::new(); - } - let mut palindrome_cache = vec![vec![None; s.len()]; s.len()]; - let mut res: Vec>> = Vec::with_capacity(s.len()); - res.push(vec![vec![(0, 1)]]); - for n in 1..s.len() { - let mut curr: Vec> = Vec::new(); - for i in 0..n + 1 { - if Solution::is_palindrome(&mut palindrome_cache, &s, i, n) { - if i > 0 { - for vec in res[i - 1].iter() { - let mut new_vec = vec.clone(); - new_vec.push((i, n + 1)); - curr.push(new_vec); - } - } else { - curr.push(vec![(i, n + 1)]); - } - } - } - res.push(curr); - } - (*res[s.len() - 1]) - .into_iter() - .map(|vec| { - vec.iter() - .map(|&range| s[range.0..range.1].iter().collect::()) - .collect::>() - }) - .collect() - } - - fn is_palindrome( - cache: &mut Vec>>, - s: &Vec, - i: usize, - j: usize, - ) -> bool { - if j <= i { - return true; - } - if let Some(result) = cache[i][j] { - result - } else { - let result = s[i] == s[j] - && (i + 1 > s.len() || j < 1 || Solution::is_palindrome(cache, s, i + 1, j - 1)); - cache[i][j] = Some(result); - result - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_131() { - assert_eq!( - Solution::partition("aab".to_owned()), - vec![vec_string!["aa", "b"], vec_string!["a", "a", "b"],] - ); - assert_eq!( - Solution::partition("aaa".to_owned()), - vec![ - vec_string!["aaa"], - vec_string!["a", "aa"], - vec_string!["aa", "a"], - vec_string!["a", "a", "a"], - ] - ); - } -} diff --git a/src/solution/s0132_palindrome_partitioning_ii.rs b/src/solution/s0132_palindrome_partitioning_ii.rs deleted file mode 100644 index 81b75cf6..00000000 --- a/src/solution/s0132_palindrome_partitioning_ii.rs +++ /dev/null @@ -1,91 +0,0 @@ -/** - * [132] Palindrome Partitioning II - * - * Given a string s, partition s such that every substring of the partition is a palindrome. - * - * Return the minimum cuts needed for a palindrome partitioning of s. - * - * Example: - * - * - * Input: "aab" - * Output: 1 - * Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/palindrome-partitioning-ii/ -// discuss: https://leetcode.com/problems/palindrome-partitioning-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -为了方便讨论, 我们记 n 个字符的最少回文分段是 f(n), 则切分次数为 f(n)-1, 接下来递推 f(n): - -f(n) = min(f(n-i) + 1) { i in [0..n] and s[i..n] is palindrome } - -显然, f(1) 为 1, f(0) 为 0 - -判断 is_palindrome 也需要优化, 使用一个备忘录, 将判断回文的操作优化到 O(1): - -is_palindrome(s[i..n]) = s[i] == s[n] && is_palindrome(s[i+1..n-1]) - -最后的复杂度: 时间 O(N^2), 空间 O(N^2) -*/ -impl Solution { - pub fn min_cut(s: String) -> i32 { - let s = s.chars().collect::>(); - if s.is_empty() { - return 0; - } - let mut palindrome_cache: Vec>> = vec![vec![None; s.len()]; s.len()]; - let mut min = Vec::with_capacity(s.len() + 1); - min.push(0); - min.push(1); - for i in 1..s.len() { - let mut local_min = i32::max_value(); - for j in 0..i + 1 { - if Solution::is_palindrome(&mut palindrome_cache, &s, j, i) { - local_min = i32::min(1 + min[j], local_min); - } - } - min.push(local_min); - } - min[s.len()] - 1 - } - - fn is_palindrome( - cache: &mut Vec>>, - s: &Vec, - i: usize, - j: usize, - ) -> bool { - if j <= i { - return true; - } - if let Some(result) = cache[i][j] { - result - } else { - let result = s[i] == s[j] - && (i + 1 > s.len() || j < 1 || Solution::is_palindrome(cache, s, i + 1, j - 1)); - cache[i][j] = Some(result); - result - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_132() { - assert_eq!(Solution::min_cut("aab".to_owned()), 1); - assert_eq!(Solution::min_cut("aaa".to_owned()), 0); - assert_eq!(Solution::min_cut("aabb".to_owned()), 1); - } -} diff --git a/src/solution/s0134_gas_station.rs b/src/solution/s0134_gas_station.rs deleted file mode 100644 index 352990f9..00000000 --- a/src/solution/s0134_gas_station.rs +++ /dev/null @@ -1,77 +0,0 @@ -/** - * [134] Gas Station - * - * There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. - * - * You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations. - * - * Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. - * - * Note: - * - * - * If there exists a solution, it is guaranteed to be unique. - * Both input arrays are non-empty and have the same length. - * Each element in the input arrays is a non-negative integer. - * - * - * Example 1: - * - * - * Input: - * gas = [1,2,3,4,5] - * cost = [3,4,5,1,2] - * - * Output: 3 - * - * Explanation: - * Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 - * Travel to station 4. Your tank = 4 - 1 + 5 = 8 - * Travel to station 0. Your tank = 8 - 2 + 1 = 7 - * Travel to station 1. Your tank = 7 - 3 + 2 = 6 - * Travel to station 2. Your tank = 6 - 4 + 3 = 5 - * Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. - * Therefore, return 3 as the starting index. - * - * - * Example 2: - * - * - * Input: - * gas = [2,3,4] - * cost = [3,4,3] - * - * Output: -1 - * - * Explanation: - * You can't start at station 0 or 1, as there is not enough gas to travel to the next station. - * Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 - * Travel to station 0. Your tank = 4 - 3 + 2 = 3 - * Travel to station 1. Your tank = 3 - 3 + 3 = 3 - * You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. - * Therefore, you can't travel around the circuit once no matter where you start. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/gas-station/ -// discuss: https://leetcode.com/problems/gas-station/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn can_complete_circuit(gas: Vec, cost: Vec) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_134() {} -} diff --git a/src/solution/s0135_candy.rs b/src/solution/s0135_candy.rs deleted file mode 100644 index debb29d0..00000000 --- a/src/solution/s0135_candy.rs +++ /dev/null @@ -1,84 +0,0 @@ -/** - * [135] Candy - * - * There are N children standing in a line. Each child is assigned a rating value. - * - * You are giving candies to these children subjected to the following requirements: - * - * - * Each child must have at least one candy. - * Children with a higher rating get more candies than their neighbors. - * - * - * What is the minimum candies you must give? - * - * Example 1: - * - * - * Input: [1,0,2] - * Output: 5 - * Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively. - * - * - * Example 2: - * - * - * Input: [1,2,2] - * Output: 4 - * Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively. - * The third child gets 1 candy because it satisfies the above two conditions. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/candy/ -// discuss: https://leetcode.com/problems/candy/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn candy(ratings: Vec) -> i32 { - let mut from = 0; - let mut n = 1; - let mut last = 1; - let mut ascending = false; - for i in 1..ratings.len() { - if ratings[i] == ratings[i - 1] { - n += 1; - from = i; - ascending = false; - } else if ratings[i] >= ratings[i - 1] { - from = i; - ascending = true; - last += 1; - n += last; - } else { - if ascending { - last = 1; - from = i; - ascending = false; - } - n += (i - from + 1) as i32; - } - } - n - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_135() { - assert_eq!(Solution::candy(vec![3, 2, 1, 2, 3]), 11); - assert_eq!(Solution::candy(vec![2, 2, 1, 2, 2]), 7); - assert_eq!(Solution::candy(vec![1, 0, 2]), 5); - assert_eq!(Solution::candy(vec![1, 2, 2]), 4); - assert_eq!(Solution::candy(vec![1, 1, 1, 1, 1, 1]), 6); - assert_eq!(Solution::candy(vec![1, 2, 2, 2, 2, 2, 2, 0]), 10); - } -} diff --git a/src/solution/s0136_single_number.rs b/src/solution/s0136_single_number.rs deleted file mode 100644 index 580d1213..00000000 --- a/src/solution/s0136_single_number.rs +++ /dev/null @@ -1,48 +0,0 @@ -/** - * [136] Single Number - * - * Given a non-empty array of integers, every element appears twice except for one. Find that single one. - * - * Note: - * - * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? - * - * Example 1: - * - * - * Input: [2,2,1] - * Output: 1 - * - * - * Example 2: - * - * - * Input: [4,1,2,1,2] - * Output: 4 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/single-number/ -// discuss: https://leetcode.com/problems/single-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -impl Solution { - pub fn single_number(nums: Vec) -> i32 { - nums.iter().fold(0, |acc, &num| acc ^ num) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_136() { - assert_eq!(Solution::single_number(vec![2, 2, 1]), 1); - assert_eq!(Solution::single_number(vec![4, 1, 2, 1, 2]), 4); - } -} diff --git a/src/solution/s0137_single_number_ii.rs b/src/solution/s0137_single_number_ii.rs deleted file mode 100644 index 5e99cd84..00000000 --- a/src/solution/s0137_single_number_ii.rs +++ /dev/null @@ -1,94 +0,0 @@ -/** - * [137] Single Number II - * - * Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one. - * - * Note: - * - * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? - * - * Example 1: - * - * - * Input: [2,2,3,2] - * Output: 3 - * - * - * Example 2: - * - * - * Input: [0,1,0,1,0,1,99] - * Output: 99 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/single-number-ii/ -// discuss: https://leetcode.com/problems/single-number-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -糅合了一下 https://leetcode.com/problems/single-number-ii/discuss/43296/An-General-Way-to-Handle-All-this-sort-of-questions. 和 https://leetcode.com/problems/single-number-ii/discuss/43294/Challenge-me-thx - -第一个链接给出了通用解法: 对于一个数出现 M 次其它数都出现了 K 的场景, 我们可以用位运算记录 K 种状态(作为一个计数器)来解 - -这题的真值表(3种状态使用2位): - -a b c/c a'b'/a'b' -0 0 1/0 0 1 /0 0 -0 1 1/0 1 0 /0 1 -1 0 1/0 0 0 /1 0 - -根据数电的知识, 要根据这个真值表写出逻辑表达式, 以输出端为 '1' 的结果为准, 将每行的输入变量写成 AND 形式, 其中为 0 的输入量需要取反, 再将这几个 AND 形式做 OR 即可 - -令 a' = 1, 则: - -a b c a' -0 1 1 1 ~a & b & c -1 0 0 1 a & ~b & ~c - -a' = (~a & b & c) | (a & ~b & ~c) - -同理: - -b' = (~a & b & ~c) | (~a & ~b & c) - -这个每轮计算的位次数达到 17 次, 可以再优化一下: - -对 b' 化简: b' = ~a & (b & ~c | ~b & c) = ~a & b ^ c - -但这时 a 仍然比较复杂, 我们可以考虑能否用每轮算出的 b' 来简化 a 的计算, 则: - -a (b) b' c a' b' -1 (0) 0 0 1 0 -0 (1) 0 1 1 0 - -重写一下就是 a' = (a & ~b' & ~c) | (~a & ~b' & c) = ~b' & (a & ~c | ~a & c) = ~b' & a ^ c - -这个就和最开始第二链接里给出的超简洁解法一致了 - -最后的话, a 或 b 为 1 都可以输出到 1 (目标数出现1次或出现2次), 输出 a | b 即可 -*/ -impl Solution { - pub fn single_number(nums: Vec) -> i32 { - let (mut a, mut b) = (0, 0); - for &num in nums.iter() { - b = !a & (b ^ num); - a = !b & (a ^ num); - } - return a | b; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_137() { - assert_eq!(Solution::single_number(vec![0, 0, 0, 1, 1, 1, 5]), 5); - } -} diff --git a/src/solution/s0139_word_break.rs b/src/solution/s0139_word_break.rs deleted file mode 100644 index ce9e2684..00000000 --- a/src/solution/s0139_word_break.rs +++ /dev/null @@ -1,93 +0,0 @@ -/** - * [139] Word Break - * - * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. - * - * Note: - * - * - * The same word in the dictionary may be reused multiple times in the segmentation. - * You may assume the dictionary does not contain duplicate words. - * - * - * Example 1: - * - * - * Input: s = "leetcode", wordDict = ["leet", "code"] - * Output: true - * Explanation: Return true because "leetcode" can be segmented as "leet code". - * - * - * Example 2: - * - * - * Input: s = "applepenapple", wordDict = ["apple", "pen"] - * Output: true - * Explanation: Return true because "applepenapple" can be segmented as "apple pen apple". - * Note that you are allowed to reuse a dictionary word. - * - * - * Example 3: - * - * - * Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] - * Output: false - * - * - */ -pub struct Solution {} - -/* -记 f[n] 表示从 0 开始长度为 n 的 substring 是否可以被组成，那么： - -f[n] = f[k] && (s[k..n] in dict) -f[0] = true - -DP 向上递推即可 - -BFS 也是可以的 -*/ - -// problem: https://leetcode.com/problems/word-break/ -// discuss: https://leetcode.com/problems/word-break/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashSet; -impl Solution { - pub fn word_break(s: String, word_dict: Vec) -> bool { - let word_dict = word_dict.into_iter().collect::>(); - let mut dp = vec![false; s.len() + 1]; - dp[0] = true; - for i in 1..s.len() + 1 { - for j in 0..s.len() { - if dp[j] && word_dict.contains(&s[j..i]) { - dp[i] = true; - } - } - } - dp[s.len()] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_139() { - assert_eq!( - Solution::word_break("leetcode".to_owned(), vec_string!["leet", "code"]), - true - ); - assert_eq!( - Solution::word_break( - "catsandog".to_owned(), - vec_string!["cats", "dog", "sand", "and", "cat"] - ), - false - ); - } -} diff --git a/src/solution/s0140_word_break_ii.rs b/src/solution/s0140_word_break_ii.rs deleted file mode 100644 index 68c1319e..00000000 --- a/src/solution/s0140_word_break_ii.rs +++ /dev/null @@ -1,72 +0,0 @@ -/** - * [140] Word Break II - * - * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. - * - * Note: - * - * - * The same word in the dictionary may be reused multiple times in the segmentation. - * You may assume the dictionary does not contain duplicate words. - * - * - * Example 1: - * - * - * Input: - * s = "catsanddog" - * wordDict = ["cat", "cats", "and", "sand", "dog"] - * Output: - * [ - * "cats and dog", - * "cat sand dog" - * ] - * - * - * Example 2: - * - * - * Input: - * s = "pineapplepenapple" - * wordDict = ["apple", "pen", "applepen", "pine", "pineapple"] - * Output: - * [ - * "pine apple pen apple", - * "pineapple pen apple", - * "pine applepen apple" - * ] - * Explanation: Note that you are allowed to reuse a dictionary word. - * - * - * Example 3: - * - * - * Input: - * s = "catsandog" - * wordDict = ["cats", "dog", "sand", "and", "cat"] - * Output: - * [] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-break-ii/ -// discuss: https://leetcode.com/problems/word-break-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn word_break(s: String, word_dict: Vec) -> Vec { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_140() {} -} diff --git a/src/solution/s0143_reorder_list.rs b/src/solution/s0143_reorder_list.rs deleted file mode 100644 index 1a7854cc..00000000 --- a/src/solution/s0143_reorder_list.rs +++ /dev/null @@ -1,50 +0,0 @@ -/** - * [143] Reorder List - * - * Given a singly linked list L: L0→L1→…→Ln-1→Ln,
- * reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… - * - * You may not modify the values in the list's nodes, only nodes itself may be changed. - * - * Example 1: - * - * - * Given 1->2->3->4, reorder it to 1->4->2->3. - * - * Example 2: - * - * - * Given 1->2->3->4->5, reorder it to 1->5->2->4->3. - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/reorder-list/ -// discuss: https://leetcode.com/problems/reorder-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* - 1->2->3->4->5 - - 1->2->3<-4<-5 - - 1->5->2->4->3 -*/ -impl Solution { - pub fn reorder_list(head: &mut Option>) { - // TODO - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_143() {} -} diff --git a/src/solution/s0144_binary_tree_preorder_traversal.rs b/src/solution/s0144_binary_tree_preorder_traversal.rs deleted file mode 100644 index 3d823976..00000000 --- a/src/solution/s0144_binary_tree_preorder_traversal.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [144] Binary Tree Preorder Traversal - * - * Given a binary tree, return the preorder traversal of its nodes' values. - * - * Example: - * - * - * Input: [1,null,2,3] - * 1 - * \ - * 2 - * / - * 3 - * - * Output: [1,2,3] - * - * - * Follow up: Recursive solution is trivial, could you do it iteratively? - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-preorder-traversal/ -// discuss: https://leetcode.com/problems/binary-tree-preorder-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn preorder_traversal(root: Option>>) -> Vec { - let mut res = Vec::new(); - Solution::helper(root, &mut res); - res - } - - fn helper(root: Option>>, vec: &mut Vec) { - if let Some(node) = root { - vec.push(node.borrow().val); - Solution::helper(node.borrow().left.clone(), vec); - Solution::helper(node.borrow().right.clone(), vec); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_144() { - assert_eq!( - Solution::preorder_traversal(tree![1, null, 2, 3]), - vec![1, 2, 3] - ); - } -} diff --git a/src/solution/s0145_binary_tree_postorder_traversal.rs b/src/solution/s0145_binary_tree_postorder_traversal.rs deleted file mode 100644 index 63197e88..00000000 --- a/src/solution/s0145_binary_tree_postorder_traversal.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [145] Binary Tree Postorder Traversal - * - * Given a binary tree, return the postorder traversal of its nodes' values. - * - * Example: - * - * - * Input: [1,null,2,3] - * 1 - * \ - * 2 - * / - * 3 - * - * Output: [3,2,1] - * - * - * Follow up: Recursive solution is trivial, could you do it iteratively? - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-postorder-traversal/ -// discuss: https://leetcode.com/problems/binary-tree-postorder-traversal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn postorder_traversal(root: Option>>) -> Vec { - let mut res = Vec::new(); - Solution::helper(root, &mut res); - res - } - - fn helper(root: Option>>, vec: &mut Vec) { - if let Some(node) = root { - Solution::helper(node.borrow().left.clone(), vec); - Solution::helper(node.borrow().right.clone(), vec); - vec.push(node.borrow().val); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_145() { - assert_eq!( - Solution::postorder_traversal(tree![1, null, 2, 3]), - vec![3, 2, 1] - ); - } -} diff --git a/src/solution/s0146_lru_cache.rs b/src/solution/s0146_lru_cache.rs deleted file mode 100644 index 226cbf77..00000000 --- a/src/solution/s0146_lru_cache.rs +++ /dev/null @@ -1,200 +0,0 @@ -/** - * [146] LRU Cache - * - * - * Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put. - * - * - * - * get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
- * put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item. - * - * - * Follow up:
- * Could you do both operations in O(1) time complexity? - * - * Example: - * - * LRUCache cache = new LRUCache( 2 /* capacity */ ); - * - * cache.put(1, 1); - * cache.put(2, 2); - * cache.get(1); // returns 1 - * cache.put(3, 3); // evicts key 2 - * cache.get(2); // returns -1 (not found) - * cache.put(4, 4); // evicts key 1 - * cache.get(1); // returns -1 (not found) - * cache.get(3); // returns 3 - * cache.get(4); // returns 4 - * - * - */ -// problem: https://leetcode.com/problems/lru-cache/ -// discuss: https://leetcode.com/problems/lru-cache/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -Least Recently Used, 最近最少使用, 关键在于追踪每一个 entry 的 age, 每次淘汰最小的那一个 key - -假如淘汰逻辑要做到 O(1) 复杂度, 我们可以引入一个链表, 每次 touch 一个值时, 就删掉它重新 push_back, 而当达到容量要驱逐时, 则 pop_front - -Rust 的链表不支持根据引用删除任意元素，也没有 LinkedHashMap，需要自己实现一个 -*/ -use std::collections::HashMap; -use std::mem; -use std::ptr; - -// Entry is either a map entry and a link-list node -pub struct LRUEntry { - key: i32, - val: i32, - prev: *mut LRUEntry, - next: *mut LRUEntry, -} - -impl LRUEntry { - pub fn new(key: i32, val: i32) -> Self { - LRUEntry { - key: key, - val: val, - prev: ptr::null_mut(), - next: ptr::null_mut(), - } - } -} - -pub struct LRUCache { - map: HashMap>, - cap: usize, - - // head and tail is dummy node of the double-linked-list - head: *mut LRUEntry, - tail: *mut LRUEntry, -} - -impl LRUCache { - pub fn new(capacity: i32) -> Self { - let capacity = capacity as usize; - let map = HashMap::with_capacity(capacity); - let cache = LRUCache { - map: map, - cap: capacity, - head: unsafe { Box::into_raw(Box::new(mem::uninitialized::())) }, - tail: unsafe { Box::into_raw(Box::new(mem::uninitialized::())) }, - }; - unsafe { - (*cache.head).next = cache.tail; - (*cache.tail).prev = cache.head; - } - - cache - } - - pub fn get(&mut self, key: i32) -> i32 { - let (ptr, val) = match self.map.get_mut(&key) { - None => (None, None), - Some(entry) => { - let ptr: *mut LRUEntry = &mut **entry; - (Some(ptr), Some(unsafe { (*entry).val })) - } - }; - - if let Some(ptr) = ptr { - self.detach(ptr); - self.push(ptr); - } - val.unwrap_or(-1) - } - - pub fn put(&mut self, key: i32, value: i32) { - let ptr = self.map.get_mut(&key).map(|entry| { - let ptr: *mut LRUEntry = &mut **entry; - ptr - }); - - match ptr { - Some(ptr) => { - // key already exist, update it - unsafe { (*ptr).val = value }; - self.detach(ptr); - self.push(ptr); - } - None => { - // insert new key, cache is full, evict - if self.map.len() == self.cap { - let mut old_entry = self.pop().unwrap(); - old_entry.key = key; - old_entry.val = value; - self.push(&mut *old_entry); - self.map.insert(key, old_entry); - } else { - let mut new_entry = Box::new(LRUEntry::new(key, value)); - self.push(&mut *new_entry); - self.map.insert(key, new_entry); - } - } - } - } - - // pop() remove the entry from map, detach the entry from head of linked-list, and return it - fn pop(&mut self) -> Option> { - let next; - unsafe { next = (*self.head).next } - // list is empty - if next == self.tail { - return None; - } - let key = unsafe { (*next).key }; - let mut old_entry = self.map.remove(&key).unwrap(); - self.detach(&mut *old_entry); - Some(old_entry) - } - - // push() pushs an entry to the tail of linked-list - fn push(&mut self, entry: *mut LRUEntry) { - unsafe { - // prev <-> tail - // prev <-> entry <-> tail - (*entry).prev = (*self.tail).prev; - (*entry).next = self.tail; - (*self.tail).prev = entry; - (*(*entry).prev).next = entry; - } - } - - // detach() remove an entry from the linked-list - fn detach(&mut self, entry: *mut LRUEntry) { - unsafe { - // prev <-> entry <-> next - // prev <-> next - (*(*entry).prev).next = (*entry).next; - (*(*entry).next).prev = (*entry).prev; - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_146() { - println!("init cache"); - let mut lru_cache = LRUCache::new(2); - lru_cache.put(1, 1); - lru_cache.put(2, 2); - println!("return 1"); - assert_eq!(lru_cache.get(1), 1); // returns 1 - println!("evict key 2"); - lru_cache.put(3, 3); // evicts key 2 - println!("return -1"); - assert_eq!(lru_cache.get(2), -1); // returns -1 (not found) - lru_cache.put(4, 4); // evicts key 1 - assert_eq!(lru_cache.get(1), -1); // returns -1 (not found) - assert_eq!(lru_cache.get(3), 3); // returns 3 - assert_eq!(lru_cache.get(4), 4); // returns 4 - } -} diff --git a/src/solution/s0147_insertion_sort_list.rs b/src/solution/s0147_insertion_sort_list.rs deleted file mode 100644 index 09aff82b..00000000 --- a/src/solution/s0147_insertion_sort_list.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [147] Insertion Sort List - * - * Sort a linked list using insertion sort. - * - *
    - *
- * - *
- * A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list.
- * With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list
- * - * - *
    - *
- * - * Algorithm of Insertion Sort: - * - *
    - * Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. - * At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. - * It repeats until no input elements remain. - *
- * - *
- * Example 1: - * - * - * Input: 4->2->1->3 - * Output: 1->2->3->4 - * - * - * Example 2: - * - * - * Input: -1->5->3->4->0 - * Output: -1->0->3->4->5 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/insertion-sort-list/ -// discuss: https://leetcode.com/problems/insertion-sort-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO; boring -impl Solution { - pub fn insertion_sort_list(head: Option>) -> Option> { - None - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_147() {} -} diff --git a/src/solution/s0148_sort_list.rs b/src/solution/s0148_sort_list.rs deleted file mode 100644 index 78b7f08c..00000000 --- a/src/solution/s0148_sort_list.rs +++ /dev/null @@ -1,125 +0,0 @@ -/** - * [148] Sort List - * - * Sort a linked list in O(n log n) time using constant space complexity. - * - * Example 1: - * - * - * Input: 4->2->1->3 - * Output: 1->2->3->4 - * - * - * Example 2: - * - * - * Input: -1->5->3->4->0 - * Output: -1->0->3->4->5 - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/sort-list/ -// discuss: https://leetcode.com/problems/sort-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -堆排序需要额外空间, 不行 - -快排: - * 不行, 单链表要快排必须同时持有两个 mut 引用, 而 rust 里这是不可能的(不知道用 unsafe 行不行) - * 不用 rust 的话应该是可行的, Lomuto-partition, 用一个慢指针记录 no_lager_than 位置 - -归并，有点慢, 每次切分要遍历找到切分点, 而且递归栈深度 O(logN) 也不算严格的 O(1) 空间 - -Rust 标准库的 std::collections::LinkedList 都没有实现 sort() 你敢信... - -这题用 rust 解对我而言真的是 Hard 级而不是 Medium 级了... - -这里也是前置知识不足, GG 了解到链表的最佳排序方式确实就是 merge-sort -*/ -impl Solution { - pub fn sort_list(mut head: Option>) -> Option> { - let mut len = 0; - let mut ptr = head.as_ref(); - while let Some(node) = ptr { - len += 1; - ptr = node.next.as_ref(); - } - Solution::merge_sort(head, len) - } - - fn merge_sort(mut head: Option>, len: i32) -> Option> { - if len < 2 { - return head; - } - let mut next = head.as_mut(); - let mut i = 1; - while i < len / 2 { - next = next.unwrap().next.as_mut(); - i += 1; - } - let mut l2 = next.unwrap().next.take(); - let mut l1 = Solution::merge_sort(head, len / 2); - let mut l2 = Solution::merge_sort(l2, len - len / 2); - Solution::merge(l1, l2) - } - - fn merge( - mut l1: Option>, - mut l2: Option>, - ) -> Option> { - let mut dummy = Some(Box::new(ListNode::new(0))); - let mut next = dummy.as_mut(); - loop { - match (l1, l2) { - (Some(mut node1), Some(mut node2)) => { - let node = if node1.val > node2.val { - // give back ownership - l2 = node2.next.take(); - l1 = Some(node1); - node2 - } else { - l1 = node1.next.take(); - l2 = Some(node2); - node1 - }; - next.as_mut().unwrap().next = Some(node); - next = next.unwrap().next.as_mut(); - } - (Some(mut node1), None) => { - next.unwrap().next = Some(node1); - break; - } - (None, Some(mut node2)) => { - next.unwrap().next = Some(node2); - break; - } - (None, None) => break, - } - } - dummy.unwrap().next - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_148() { - assert_eq!( - Solution::sort_list(linked![4, 2, 1, 3]), - linked![1, 2, 3, 4] - ); - assert_eq!( - Solution::sort_list(linked![-1, 5, 3, 4, 0]), - linked![-1, 0, 3, 4, 5] - ); - assert_eq!(Solution::sort_list(linked![]), linked![]); - } -} diff --git a/src/solution/s0149_max_points_on_a_line.rs b/src/solution/s0149_max_points_on_a_line.rs deleted file mode 100644 index 31ea4c62..00000000 --- a/src/solution/s0149_max_points_on_a_line.rs +++ /dev/null @@ -1,178 +0,0 @@ -/** - * [149] Max Points on a Line - * - * Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. - * - * Example 1: - * - * - * Input: [[1,1],[2,2],[3,3]] - * Output: 3 - * Explanation: - * ^ - * | - * | o - * | o - * | o - * +-------------> - * 0 1 2 3 4 - * - * - * Example 2: - * - * - * Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] - * Output: 4 - * Explanation: - * ^ - * | - * | o - * | o o - * | o - * | o o - * +-------------------> - * 0 1 2 3 4 5 6 - * - * - */ -pub struct Solution {} -use crate::util::point::Point; - -// problem: https://leetcode.com/problems/max-points-on-a-line/ -// discuss: https://leetcode.com/problems/max-points-on-a-line/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -要回顾下高中数学：已知两点, 求解一般式: - - * Ax + By + C = 0 - * A = y2 - y1, B = x1 - x2, C = x2y1 - x1y2 - -有这个知识之后，化为一般式，做三层遍历就行，再加上一个 HashSet，避免对同一直线上点的重复计算，时间复杂度可以是 O(N^2) - -有两个坑要注意避免： - - * 给的 case 会导致 i32 溢出，这里直接用了 i64 表示 - * 给的 case 里有相同的点，直接处理相同点的话会导致最坏情况复杂度到 O(N^3)，因此要先做一次转化，归并相同的点 - -用 Rust 实现有另一点注意的，给的 Point 没有实现 Hash Trait，要自己转化一下 -*/ -// straight-line expression: Ax + By + C = 0 -// A = y2 - y1, B = x1 - x2, C = x2y1 - x1y2 -#[derive(PartialEq, Hash, Eq, Debug)] -struct Line(i64, i64, i64); - -impl Line { - // Assumes that there is no same point - fn new(p1: &Point, p2: &Point) -> Self { - let x1 = p1.x as i64; - let x2 = p2.x as i64; - let y1 = p1.y as i64; - let y2 = p2.y as i64; - Line(y2 - y1, x1 - x2, x2 * y1 - x1 * y2) - } - fn contains(&self, p: &Point) -> bool { - self.0 * p.x as i64 + self.1 * p.y as i64 + self.2 == 0_i64 - } -} - -use std::collections::HashMap; -use std::collections::HashSet; -impl Solution { - pub fn max_points(points: Vec) -> i32 { - // fold same point, record the point count - let points: Vec<(Point, i32)> = points - .into_iter() - .fold(HashMap::new(), |mut map, v| { - *map.entry((v.x, v.y)).or_insert(0) += 1; - map - }) - .into_iter() - .map(|(k, v)| (Point::new(k.0, k.1), v)) // Point did not implement Hash trait - .collect(); - - // any two points in a straight-line, return quickly - if points.len() < 3 { - return points.into_iter().fold(0, |acc, v| acc + v.1); - } - let mut max = 2; - let mut set: HashSet = HashSet::new(); - for i in 0..(points.len() - 1) { - for j in i + 1..points.len() { - let line = Line::new(&points[i].0, &points[j].0); - if set.contains(&line) { - continue; - } - let mut curr = points[i].1 + points[j].1; - for k in j + 1..points.len() { - if line.contains(&points[k].0) { - curr += points[k].1; - } - } - max = i32::max(max, curr); - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_149() { - assert_eq!( - Solution::max_points(vec![point![1, 1], point![2, 2], point![3, 3]]), - 3 - ); - assert_eq!( - Solution::max_points(vec![ - point![1, 1], - point![3, 2], - point![5, 3], - point![4, 1], - point![2, 3], - point![1, 4] - ]), - 4 - ); - assert_eq!( - Solution::max_points(vec![point![0, 0], point![1, 65536], point![65536, 0]]), - 2 - ); - assert_eq!( - Solution::max_points(vec![point![1, 1], point![1, 1], point![1, 1]]), - 3 - ); - assert_eq!( - Solution::max_points(vec![ - point![0, 9], - point![138, 429], - point![115, 359], - point![115, 359], - point![-30, -102], - point![230, 709], - point![-150, -686], - point![-135, -613], - point![-60, -248], - point![-161, -481], - point![207, 639], - point![23, 79], - point![-230, -691], - point![-115, -341], - point![92, 289], - point![60, 336], - point![-105, -467], - point![135, 701], - point![-90, -394], - point![-184, -551], - point![150, 774] - ]), - 12 - ) - } -} diff --git a/src/solution/s0150_evaluate_reverse_polish_notation.rs b/src/solution/s0150_evaluate_reverse_polish_notation.rs deleted file mode 100644 index 048824dd..00000000 --- a/src/solution/s0150_evaluate_reverse_polish_notation.rs +++ /dev/null @@ -1,91 +0,0 @@ -/** - * [150] Evaluate Reverse Polish Notation - * - * Evaluate the value of an arithmetic expression in Reverse Polish Notation. - * - * Valid operators are +, -, *, /. Each operand may be an integer or another expression. - * - * Note: - * - * - * Division between two integers should truncate toward zero. - * The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation. - * - * - * Example 1: - * - * - * Input: ["2", "1", "+", "3", "*"] - * Output: 9 - * Explanation: ((2 + 1) * 3) = 9 - * - * - * Example 2: - * - * - * Input: ["4", "13", "5", "/", "+"] - * Output: 6 - * Explanation: (4 + (13 / 5)) = 6 - * - * - * Example 3: - * - * - * Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] - * Output: 22 - * Explanation: - * ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 - * = ((10 * (6 / (12 * -11))) + 17) + 5 - * = ((10 * (6 / -132)) + 17) + 5 - * = ((10 * 0) + 17) + 5 - * = (0 + 17) + 5 - * = 17 + 5 - * = 22 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/evaluate-reverse-polish-notation/ -// discuss: https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn eval_rpn(tokens: Vec) -> i32 { - let mut stack: Vec = Vec::new(); - for t in tokens.iter() { - if let Ok(num) = t.parse::() { - stack.push(num); - } else { - let right = stack.pop().unwrap(); - let left = stack.pop().unwrap(); - match (t as &str) { - "*" => stack.push(left * right), - "+" => stack.push(left + right), - "/" => stack.push(left / right), - "-" => stack.push(left - right), - _ => unreachable!(), - } - } - } - stack.pop().unwrap() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_150() { - assert_eq!( - Solution::eval_rpn(vec_string![ - "10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+" - ]), - 22 - ); - } -} diff --git a/src/solution/s0151_reverse_words_in_a_string.rs b/src/solution/s0151_reverse_words_in_a_string.rs deleted file mode 100644 index 627f0dd6..00000000 --- a/src/solution/s0151_reverse_words_in_a_string.rs +++ /dev/null @@ -1,93 +0,0 @@ -/** - * [151] Reverse Words in a String - * - * Given an input string, reverse the string word by word. - * - * - * - * Example 1: - * - * - * Input: "the sky is blue" - * Output: "blue is sky the" - * - * - * Example 2: - * - * - * Input: " hello world! " - * Output: "world! hello" - * Explanation: Your reversed string should not contain leading or trailing spaces. - * - * - * Example 3: - * - * - * Input: "a good example" - * Output: "example good a" - * Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string. - * - * - * - * - * Note: - * - * - * A word is defined as a sequence of non-space characters. - * Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces. - * You need to reduce multiple spaces between two words to a single space in the reversed string. - * - * - * - * - * Follow up: - * - * For C programmers, try to solve it in-place in O(1) extra space. - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/reverse-words-in-a-string/ -// discuss: https://leetcode.com/problems/reverse-words-in-a-string/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn reverse_words(mut s: String) -> String { - let mut seq = s.trim().chars().collect::>(); - seq.reverse(); - let mut start_idx = 0_usize; - let mut i = 0_usize; - while i < seq.len() { - if seq[i] == ' ' { - if i == start_idx { - seq.remove(i); - continue; - } - seq[start_idx..i].reverse(); - start_idx = i + 1; - } - i += 1; - } - seq[start_idx..].reverse(); - seq.into_iter().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_151() { - assert_eq!( - Solution::reverse_words("the sky is blue".to_owned()), - "blue is sky the".to_owned() - ); - assert_eq!( - Solution::reverse_words(" hello world! ".to_owned()), - "world! hello".to_owned() - ); - } -} diff --git a/src/solution/s0152_maximum_product_subarray.rs b/src/solution/s0152_maximum_product_subarray.rs deleted file mode 100644 index b85b5ecb..00000000 --- a/src/solution/s0152_maximum_product_subarray.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [152] Maximum Product Subarray - * - * Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product. - * - * Example 1: - * - * - * Input: [2,3,-2,4] - * Output: 6 - * Explanation: [2,3] has the largest product 6. - * - * - * Example 2: - * - * - * Input: [-2,0,-1] - * Output: 0 - * Explanation: The result cannot be 2, because [-2,-1] is not a subarray. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/maximum-product-subarray/ -// discuss: https://leetcode.com/problems/maximum-product-subarray/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -f[i], g[i] means the max positive value and max negative value for the sub-seq end with index i - -then we have: - -f[i], g[i] = if nums[i] == 0 { - 0, 0 -} else if nums[i] > 0 { - f[i-1] * nums[i], g[i-1] * nums[i] -} else if nums[i] < 0 { - g[i-1] * nums[i], f[i-1] * nums[i] -} -*/ - -impl Solution { - pub fn max_product(nums: Vec) -> i32 { - let mut max = nums[0]; - let mut neg_max = 0; - let mut pos_max = 0; - for num in nums.into_iter() { - if num == 0 { - neg_max = 0; - pos_max = 0; - max = i32::max(max, 0); - } else if num > 0 { - pos_max = i32::max(pos_max * num, num); - neg_max = neg_max * num; - } else { - let pos_pre = pos_max; - pos_max = neg_max * num; - neg_max = i32::min(pos_pre * num, num); - } - if pos_max != 0 { - max = i32::max(max, pos_max); - } - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_152() { - assert_eq!(Solution::max_product(vec![2, 3, -2, 4]), 6); - assert_eq!(Solution::max_product(vec![-2, 0, -1]), 0); - assert_eq!(Solution::max_product(vec![-4, -3, -2]), 12); - } -} diff --git a/src/solution/s0153_find_minimum_in_rotated_sorted_array.rs b/src/solution/s0153_find_minimum_in_rotated_sorted_array.rs deleted file mode 100644 index 03b7e7f1..00000000 --- a/src/solution/s0153_find_minimum_in_rotated_sorted_array.rs +++ /dev/null @@ -1,64 +0,0 @@ -/** - * [153] Find Minimum in Rotated Sorted Array - * - * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. - * - * (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). - * - * Find the minimum element. - * - * You may assume no duplicate exists in the array. - * - * Example 1: - * - * - * Input: [3,4,5,1,2] - * Output: 1 - * - * - * Example 2: - * - * - * Input: [4,5,6,7,0,1,2] - * Output: 0 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ -// discuss: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn find_min(nums: Vec) -> i32 { - let mut size = nums.len(); - if size == 0 { - return -1; - } - let mut base = 0_usize; - while size > 1 { - let half = size / 2; - let mid = base + half; - if nums[mid] > nums[base] { - base = mid; - } - size -= half; - } - i32::min(nums[base], nums[(base + 1) % nums.len()]) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_153() { - assert_eq!(Solution::find_min(vec![4, 5, 6, 1, 2, 3]), 1); - assert_eq!(Solution::find_min(vec![4, 5, 6, 7, 0, 1, 2]), 0); - } -} diff --git a/src/solution/s0154_find_minimum_in_rotated_sorted_array_ii.rs b/src/solution/s0154_find_minimum_in_rotated_sorted_array_ii.rs deleted file mode 100644 index cdb23ac3..00000000 --- a/src/solution/s0154_find_minimum_in_rotated_sorted_array_ii.rs +++ /dev/null @@ -1,79 +0,0 @@ -/** - * [154] Find Minimum in Rotated Sorted Array II - * - * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. - * - * (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). - * - * Find the minimum element. - * - * The array may contain duplicates. - * - * Example 1: - * - * - * Input: [1,3,5] - * Output: 1 - * - * Example 2: - * - * - * Input: [2,2,2,0,1] - * Output: 0 - * - * Note: - * - * - * This is a follow up problem to Find Minimum in Rotated Sorted Array. - * Would allow duplicates affect the run-time complexity? How and why? - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ -// discuss: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -针对无重复的做法, 只要二分搜索找折点即可: 假如 nums[mid] > nums[base] 那么转折点一定在右侧, 否则在左侧 - -但假如有重复, 就可能有 nums[mid] == nums[base], 这时就尴尬了, 无法确定转折点在左半部分还是右半部分 - -可以考虑一个数组, [1,1,1,1,1,1,1,0,1,1,1,1,1,1] 这个数组无论怎么去找 0, 时间复杂度无法低于 O(N) - -但假如不是这种极端情况, 那么二分搜索还是能优化的, 在 153 的基础上, 碰到相等就跳过即可 -*/ -impl Solution { - pub fn find_min(nums: Vec) -> i32 { - let mut lo = 0; - let mut hi = nums.len() - 1; - let mut mid = 0; - while lo < hi { - mid = lo + (hi - lo) / 2; - if (nums[mid] > nums[hi]) { - lo = mid + 1; - } else if (nums[mid] < nums[hi]) { - hi = mid; - } else { - hi -= 1; - } - } - return nums[lo]; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_154() { - assert_eq!(Solution::find_min(vec![1, 2, 2, 2, 2, 2]), 1); - assert_eq!(Solution::find_min(vec![1, 3, 3]), 1); - assert_eq!(Solution::find_min(vec![3, 1, 3, 3]), 1); - } -} diff --git a/src/solution/s0155_min_stack.rs b/src/solution/s0155_min_stack.rs deleted file mode 100644 index 99a96698..00000000 --- a/src/solution/s0155_min_stack.rs +++ /dev/null @@ -1,116 +0,0 @@ -/** - * [155] Min Stack - * - * - * Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. - * - * - * push(x) -- Push element x onto stack. - * - * - * pop() -- Removes the element on top of the stack. - * - * - * top() -- Get the top element. - * - * - * getMin() -- Retrieve the minimum element in the stack. - * - * - * - * - * Example:
- * - * MinStack minStack = new MinStack(); - * minStack.push(-2); - * minStack.push(0); - * minStack.push(-3); - * minStack.getMin(); --> Returns -3. - * minStack.pop(); - * minStack.top(); --> Returns 0. - * minStack.getMin(); --> Returns -2. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/min-stack/ -// discuss: https://leetcode.com/problems/min-stack/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -这题居然是 easy... 我怀疑人生了, getMin() 怎么能做到常数时间? Heap 也是 LogN 啊 - -看了最高票解之后...........天哪, 我可太菜了 - -核心思想是保证每次 pop 时都能以常数时间更新最小值, 这就需要在空间上以某种方式记录下来 - -那一种做法就是存储每个元素和最小值之间的差值, 这样 pop 的时候就能不断还原出原始值 - -另一种更直观的做法就是每次入栈 min 时, 都把前一个 min (当前第二小的数字) 放在它前面, 作为记录 -*/ -struct MinStack { - vec: Vec, - min: i32, -} - -impl MinStack { - /** initialize your data structure here. */ - pub fn new() -> Self { - MinStack { - vec: Vec::new(), - min: i32::max_value(), - } - } - - pub fn push(&mut self, x: i32) { - if x <= self.min { - self.vec.push(self.min); - self.min = x; - } - self.vec.push(x); - } - - pub fn pop(&mut self) { - if self.vec.pop().unwrap() == self.min { - self.min = self.vec.pop().unwrap(); - } - } - - pub fn top(&self) -> i32 { - *self.vec.last().unwrap() - } - - pub fn get_min(&self) -> i32 { - self.min - } -} - -/** - * Your MinStack object will be instantiated and called as such: - * let obj = MinStack::new(); - * obj.push(x); - * obj.pop(); - * let ret_3: i32 = obj.top(); - * let ret_4: i32 = obj.get_min(); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - pub fn test_155() { - let mut min_stack = MinStack::new(); - min_stack.push(-2); - min_stack.push(0); - min_stack.push(-3); - assert_eq!(min_stack.get_min(), -3); // --> Returns -3. - min_stack.pop(); - assert_eq!(min_stack.top(), 0); // --> Returns 0. - assert_eq!(min_stack.get_min(), -2); // --> Returns -2.[] - } -} diff --git a/src/solution/s0162_find_peak_element.rs b/src/solution/s0162_find_peak_element.rs deleted file mode 100644 index 3888b998..00000000 --- a/src/solution/s0162_find_peak_element.rs +++ /dev/null @@ -1,67 +0,0 @@ -/** - * [162] Find Peak Element - * - * A peak element is an element that is greater than its neighbors. - * - * Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. - * - * The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. - * - * You may imagine that nums[-1] = nums[n] = -∞. - * - * Example 1: - * - * - * Input: nums = [1,2,3,1] - * Output: 2 - * Explanation: 3 is a peak element and your function should return the index number 2. - * - * Example 2: - * - * - * Input: nums = [1,2,1,3,5,6,4] - * Output: 1 or 5 - * Explanation: Your function can return either index number 1 where the peak element is 2, - * or index number 5 where the peak element is 6. - * - * - * Note: - * - * Your solution should be in logarithmic complexity. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-peak-element/ -// discuss: https://leetcode.com/problems/find-peak-element/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn find_peak_element(nums: Vec) -> i32 { - let (mut lo, mut hi) = (0_usize, nums.len() - 1); - let mut mid = 0; - while lo < hi { - mid = (hi - lo) / 2 + lo; - if mid + 1 < nums.len() && nums[mid] < nums[mid + 1] { - lo = mid + 1; - } else { - hi = mid; - } - } - lo as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_162() { - assert_eq!(Solution::find_peak_element(vec![1, 2, 3, 1]), 2); - assert_eq!(Solution::find_peak_element(vec![1, 2, 1, 3, 5, 6, 4]), 5); - } -} diff --git a/src/solution/s0164_maximum_gap.rs b/src/solution/s0164_maximum_gap.rs deleted file mode 100644 index bd961db7..00000000 --- a/src/solution/s0164_maximum_gap.rs +++ /dev/null @@ -1,66 +0,0 @@ -/** - * [164] Maximum Gap - * - * Given an unsorted array, find the maximum difference between the successive elements in its sorted form. - * - * Return 0 if the array contains less than 2 elements. - * - * Example 1: - * - * - * Input: [3,6,9,1] - * Output: 3 - * Explanation: The sorted form of the array is [1,3,6,9], either - * (3,6) or (6,9) has the maximum difference 3. - * - * Example 2: - * - * - * Input: [10] - * Output: 0 - * Explanation: The array contains less than 2 elements, therefore return 0. - * - * Note: - * - * - * You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range. - * Try to solve it in linear time/space. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/maximum-gap/ -// discuss: https://leetcode.com/problems/maximum-gap/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -/* -想不出来, 一看解析居然是 Radix Sort 或 Bucket Sort, 我就 ??? 了... - -最佳算法是 Bucket Sort 吗? (桶大小取 max - min / len 那种), 看时间复杂度好像是这样 - -但假如整体排布非常稠密, 那么这个聪明的算法也就退化成了桶大小为 1 的桶排序 -*/ -impl Solution { - pub fn maximum_gap(nums: Vec) -> i32 { - let mut nums = nums; - nums.sort_unstable(); - let mut gap = 0; - for i in 1..nums.len() { - gap = i32::max(nums[i] - nums[i - 1], gap); - } - gap - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_164() { - assert_eq!(Solution::maximum_gap(vec![3, 6, 9, 1]), 3); - } -} diff --git a/src/solution/s0165_compare_version_numbers.rs b/src/solution/s0165_compare_version_numbers.rs deleted file mode 100644 index b53f1bbb..00000000 --- a/src/solution/s0165_compare_version_numbers.rs +++ /dev/null @@ -1,112 +0,0 @@ -/** - * [165] Compare Version Numbers - * - * Compare two version numbers version1 and version2.
- * If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0. - * - * You may assume that the version strings are non-empty and contain only digits and the . character. - * The . character does not represent a decimal point and is used to separate number sequences. - * For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision. - * You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0. - * - * - * - * Example 1: - * - * Input: version1 = "0.1", version2 = "1.1" - * Output: -1 - * - * Example 2: - * - * Input: version1 = "1.0.1", version2 = "1" - * Output: 1 - * - * Example 3: - * - * Input: version1 = "7.5.2.4", version2 = "7.5.3" - * Output: -1 - * - * Example 4: - * - * Input: version1 = "1.01", version2 = "1.001" - * Output: 0 - * Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1” - * - * Example 5: - * - * Input: version1 = "1.0", version2 = "1.0.0" - * Output: 0 - * Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0" - * - * - * - * Note: - *
    - * Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes. - * Version strings do not start or end with dots, and they will not be two consecutive dots. - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/compare-version-numbers/ -// discuss: https://leetcode.com/problems/compare-version-numbers/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn compare_version(version1: String, version2: String) -> i32 { - let v1: Vec<&str> = version1.split('.').collect::>(); - let v2: Vec<&str> = version2.split('.').collect::>(); - let mut i = 0_usize; - while i < v1.len() && i < v2.len() { - let left = v1[i].parse::().unwrap(); - let right = v2[i].parse::().unwrap(); - if left > right { - return 1; - } else if left < right { - return -1; - } - i += 1; - } - while i < v1.len() { - if v1[i].parse::().unwrap() > 0 { - return 1; - } - i += 1; - } - while i < v2.len() { - if v2[i].parse::().unwrap() > 0 { - return -1; - } - i += 1; - } - return 0; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_165() { - assert_eq!( - Solution::compare_version("0.1".to_owned(), "1.1".to_owned()), - -1 - ); - assert_eq!( - Solution::compare_version("1.0.1".to_owned(), "1".to_owned()), - 1 - ); - assert_eq!( - Solution::compare_version("7.5.2.4".to_owned(), "7.5.3".to_owned()), - -1 - ); - assert_eq!( - Solution::compare_version("1.01".to_owned(), "1.0001".to_owned()), - 0 - ); - } -} diff --git a/src/solution/s0166_fraction_to_recurring_decimal.rs b/src/solution/s0166_fraction_to_recurring_decimal.rs deleted file mode 100644 index 4585738f..00000000 --- a/src/solution/s0166_fraction_to_recurring_decimal.rs +++ /dev/null @@ -1,51 +0,0 @@ -/** - * [166] Fraction to Recurring Decimal - * - * Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. - * - * If the fractional part is repeating, enclose the repeating part in parentheses. - * - * Example 1: - * - * - * Input: numerator = 1, denominator = 2 - * Output: "0.5" - * - * - * Example 2: - * - * - * Input: numerator = 2, denominator = 1 - * Output: "2" - * - * Example 3: - * - * - * Input: numerator = 2, denominator = 3 - * Output: "0.(6)" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/fraction-to-recurring-decimal/ -// discuss: https://leetcode.com/problems/fraction-to-recurring-decimal/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO -impl Solution { - pub fn fraction_to_decimal(numerator: i32, denominator: i32) -> String { - "".to_owned() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_166() {} -} diff --git a/src/solution/s0167_two_sum_ii_input_array_is_sorted.rs b/src/solution/s0167_two_sum_ii_input_array_is_sorted.rs deleted file mode 100644 index 58de2e86..00000000 --- a/src/solution/s0167_two_sum_ii_input_array_is_sorted.rs +++ /dev/null @@ -1,58 +0,0 @@ -/** - * [167] Two Sum II - Input array is sorted - * - * Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. - * - * The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. - * - * Note: - * - * - * Your returned answers (both index1 and index2) are not zero-based. - * You may assume that each input would have exactly one solution and you may not use the same element twice. - * - * - * Example: - * - * - * Input: numbers = [2,7,11,15], target = 9 - * Output: [1,2] - * Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/ -// discuss: https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn two_sum(numbers: Vec, target: i32) -> Vec { - let mut i = 0_usize; - let mut j = numbers.len() - 1; - while i < j { - let sum = numbers[i] + numbers[j]; - if sum > target { - j -= 1; - } else if sum < target { - i += 1; - } else { - break; - } - } - return vec![i as i32 + 1, j as i32 + 1]; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_167() { - assert_eq!(Solution::two_sum(vec![2, 7, 11, 15], 9), vec![1, 2]); - } -} diff --git a/src/solution/s0168_excel_sheet_column_title.rs b/src/solution/s0168_excel_sheet_column_title.rs deleted file mode 100644 index 5a9dc585..00000000 --- a/src/solution/s0168_excel_sheet_column_title.rs +++ /dev/null @@ -1,78 +0,0 @@ -/** - * [168] Excel Sheet Column Title - * - * Given a positive integer, return its corresponding column title as appear in an Excel sheet. - * - * For example: - * - * - * 1 -> A - * 2 -> B - * 3 -> C - * ... - * 26 -> Z - * 27 -> AA - * 28 -> AB - * ... - * - * - * Example 1: - * - * - * Input: 1 - * Output: "A" - * - * - * Example 2: - * - * - * Input: 28 - * Output: "AB" - * - * - * Example 3: - * - * - * Input: 701 - * Output: "ZY" - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/excel-sheet-column-title/ -// discuss: https://leetcode.com/problems/excel-sheet-column-title/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn convert_to_title(n: i32) -> String { - let base = 26; - let mut n = n; - let mut res = Vec::new(); - while n > 0 { - let mut code = (n % base) as u8; - n = n / base; - if code == 0 { - n -= 1; - code = base as u8; - }; - let alphabetic = (('A' as u8) + (code - 1_u8)) as char; - res.push(alphabetic); - } - res.reverse(); - res.into_iter().collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_168() { - assert_eq!(Solution::convert_to_title(28), "AB".to_owned()); - assert_eq!(Solution::convert_to_title(1), "A".to_owned()); - } -} diff --git a/src/solution/s0169_majority_element.rs b/src/solution/s0169_majority_element.rs deleted file mode 100644 index be59bea6..00000000 --- a/src/solution/s0169_majority_element.rs +++ /dev/null @@ -1,73 +0,0 @@ -/** - * [169] Majority Element - * - * Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. - * - * You may assume that the array is non-empty and the majority element always exist in the array. - * - * Example 1: - * - * - * Input: [3,2,3] - * Output: 3 - * - * Example 2: - * - * - * Input: [2,2,1,1,1,2,2] - * Output: 2 - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/majority-element/ -// discuss: https://leetcode.com/problems/majority-element/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -抄的题解：Boyer-Moore Voting Algorithm -自己只能想到 HashMap 和排序, 真是太鸡儿菜了... - -Boyer-Moore Voting Algorithm 的思路是假设当前值为主元素, 碰到当前值则 +1, 非当前值则 -1, 计数器一旦归零, -就取下一个数为主元素 - -最后留下的数一定主元素 - -证明也很简单, 假设我们从第 i 位开始选择了一个数 A, 并且这个数 A 保持到了循环终止, 那么: - -我们知道, 第 nums[i..n] 中, A 是主元素, nums[0..i] 中, 有一个数 B 出现了一半的次数 - -假如 A = B, 那么 A 出现了大于一半的次数, A 一定是主元素 - -假如 A != B, 且主元素不是 A, 那么 B 包括其他任何数在整个数组中出现的次数一定不到一半(因为 B 包括其他任何数 -在前半部分**至多**出现一半, 而在后半部分不到一半), 因此不存在主元素, 这与题目给定的"一定存在主元素"矛盾, 因此 -A 一定是主元素 -*/ - -impl Solution { - pub fn majority_element(nums: Vec) -> i32 { - let mut count = 0; - let mut candidate = 0; - for &num in nums.iter() { - if count == 0 { - candidate = num; - } - count += if num == candidate { 1 } else { -1 }; - } - candidate - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_169() { - assert_eq!(Solution::majority_element(vec![2, 2, 1, 1, 1, 2, 2]), 2); - } -} diff --git a/src/solution/s0171_excel_sheet_column_number.rs b/src/solution/s0171_excel_sheet_column_number.rs deleted file mode 100644 index 1b2f611f..00000000 --- a/src/solution/s0171_excel_sheet_column_number.rs +++ /dev/null @@ -1,62 +0,0 @@ -/** - * [171] Excel Sheet Column Number - * - * Given a column title as appear in an Excel sheet, return its corresponding column number. - * - * For example: - * - * - * A -> 1 - * B -> 2 - * C -> 3 - * ... - * Z -> 26 - * AA -> 27 - * AB -> 28 - * ... - * - * - * Example 1: - * - * - * Input: "A" - * Output: 1 - * - * - * Example 2: - * - * - * Input: "AB" - * Output: 28 - * - * - * Example 3: - * - * - * Input: "ZY" - * Output: 701 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/excel-sheet-column-number/ -// discuss: https://leetcode.com/problems/excel-sheet-column-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// TODO: boring -impl Solution { - pub fn title_to_number(s: String) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_171() {} -} diff --git a/src/solution/s0172_factorial_trailing_zeroes.rs b/src/solution/s0172_factorial_trailing_zeroes.rs deleted file mode 100644 index 9a54916d..00000000 --- a/src/solution/s0172_factorial_trailing_zeroes.rs +++ /dev/null @@ -1,55 +0,0 @@ -/** - * [172] Factorial Trailing Zeroes - * - * Given an integer n, return the number of trailing zeroes in n!. - * - * Example 1: - * - * - * Input: 3 - * Output: 0 - * Explanation: 3! = 6, no trailing zero. - * - * Example 2: - * - * - * Input: 5 - * Output: 1 - * Explanation: 5! = 120, one trailing zero. - * - * Note: Your solution should be in logarithmic time complexity. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/factorial-trailing-zeroes/ -// discuss: https://leetcode.com/problems/factorial-trailing-zeroes/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn trailing_zeroes(n: i32) -> i32 { - let mut five_count = 0; - let mut n = n; - while n > 0 { - n = n / 5; - five_count += n; - } - five_count - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_172() { - assert_eq!(Solution::trailing_zeroes(3), 0); - assert_eq!(Solution::trailing_zeroes(5), 1); - assert_eq!(Solution::trailing_zeroes(20), 4); - assert_eq!(Solution::trailing_zeroes(1808548329), 452137076); - } -} diff --git a/src/solution/s0173_binary_search_tree_iterator.rs b/src/solution/s0173_binary_search_tree_iterator.rs deleted file mode 100644 index b350600d..00000000 --- a/src/solution/s0173_binary_search_tree_iterator.rs +++ /dev/null @@ -1,116 +0,0 @@ -/** - * [173] Binary Search Tree Iterator - * - * Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. - * - * Calling next() will return the next smallest number in the BST. - * - * - * - * - * - * - * Example: - * - * - * - * - * BSTIterator iterator = new BSTIterator(root); - * iterator.next(); // return 3 - * iterator.next(); // return 7 - * iterator.hasNext(); // return true - * iterator.next(); // return 9 - * iterator.hasNext(); // return true - * iterator.next(); // return 15 - * iterator.hasNext(); // return true - * iterator.next(); // return 20 - * iterator.hasNext(); // return false - * - * - * - * - * Note: - * - * - * next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree. - * You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called. - * - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; -use std::cell::RefCell; -use std::rc::Rc; - -// problem: https://leetcode.com/problems/binary-search-tree-iterator/ -// discuss: https://leetcode.com/problems/binary-search-tree-iterator/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -非递归中序遍历 -*/ -pub struct BSTIterator { - stack: Vec>>, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl BSTIterator { - pub fn new(root: Option>>) -> Self { - let mut node = root; - let mut stack = Vec::new(); - while let Some(inner) = node.clone() { - stack.push(inner.clone()); - node = node.unwrap().borrow().left.clone(); - } - BSTIterator { stack: stack } - } - - /** @return the next smallest number */ - pub fn next(&mut self) -> i32 { - let node = self.stack.pop().unwrap(); - let res = node.borrow().val; - let mut next = node.borrow().right.clone(); - while let Some(inner) = next.clone() { - self.stack.push(inner.clone()); - next = next.unwrap().borrow().left.clone(); - } - res - } - - /** @return whether we have a next smallest number */ - pub fn has_next(&self) -> bool { - !self.stack.is_empty() - } -} - -/** - * Your BSTIterator object will be instantiated and called as such: - * let obj = BSTIterator::new(root); - * let ret_1: i32 = obj.next(); - * let ret_2: bool = obj.has_next(); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_173() { - let mut iterator = BSTIterator::new(tree![7, 3, 15, null, null, 9, 20]); - assert_eq!(iterator.next(), 3); // return 3 - assert_eq!(iterator.next(), 7); // return 7 - assert_eq!(iterator.has_next(), true); // return true - assert_eq!(iterator.next(), 9); // return 9 - assert_eq!(iterator.has_next(), true); // return true - assert_eq!(iterator.next(), 15); // return 15 - assert_eq!(iterator.has_next(), true); // return true - assert_eq!(iterator.next(), 20); // return 20 - assert_eq!(iterator.has_next(), false); // return false - } -} diff --git a/src/solution/s0174_dungeon_game.rs b/src/solution/s0174_dungeon_game.rs deleted file mode 100644 index 23be4d97..00000000 --- a/src/solution/s0174_dungeon_game.rs +++ /dev/null @@ -1,115 +0,0 @@ -/** - * [174] Dungeon Game - * - * - * The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess. - * - * The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately. - * - * Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers). - * - * In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step. - * - * - * - * Write a function to determine the knight's minimum initial health so that he is able to rescue the princess. - * - * For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN. - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - *
-2 (K)-33
-5-101
1030-5 (P)
- * - * - * - * Note: - * - * - * The knight's health has no upper bound. - * Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/dungeon-game/ -// discuss: https://leetcode.com/problems/dungeon-game/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -DP, 即从每个格子出发到达终点所需的最小生命值为 hp[i][j] - -则显然, hp[M-1][N-1] = min(dungeon[M-1][N-1], 0) + 1; - -hp[i][j] = min(min(hp[i+1][j], hp[i][j+1]) - dungeon[i][j], 1); - -倒推到 hp[0][0] 即可 - -这里倒推很重要, 因为正推很难 dp(有后效性) - -其实可以优化成 O(M+N) 空间复杂度, 从斜对角线往后推就只需要保存一个小数组, 但是下面这样更简明 -*/ -impl Solution { - pub fn calculate_minimum_hp(dungeon: Vec>) -> i32 { - let (height, width) = (dungeon.len(), dungeon[0].len()); - // Using dummy row to simplify logic - let mut hp = vec![vec![i32::max_value(); width + 1]; height + 1]; - hp[height][width - 1] = 1; - hp[height - 1][width] = 1; - for i in (0..height).rev() { - for j in (0..width).rev() { - hp[i][j] = i32::max(i32::min(hp[i + 1][j], hp[i][j + 1]) - dungeon[i][j], 1); - } - } - hp[0][0] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_174() { - assert_eq!( - Solution::calculate_minimum_hp(vec![ - vec![-2, -3, 3], - vec![-5, -10, 1], - vec![10, 30, -5], - ]), - 7 - ); - assert_eq!( - Solution::calculate_minimum_hp(vec![vec![1, -4, 5, -99], vec![2, -2, -2, -1]]), - 3 - ); - } -} diff --git a/src/solution/s0179_largest_number.rs b/src/solution/s0179_largest_number.rs deleted file mode 100644 index 65348165..00000000 --- a/src/solution/s0179_largest_number.rs +++ /dev/null @@ -1,60 +0,0 @@ -/** - * [179] Largest Number - * - * Given a list of non negative integers, arrange them such that they form the largest number. - * - * Example 1: - * - * - * Input: [10,2] - * Output: "210" - * - * Example 2: - * - * - * Input: [3,30,34,5,9] - * Output: "9534330" - * - * - * Note: The result may be very large, so you need to return a string instead of an integer. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/largest-number/ -// discuss: https://leetcode.com/problems/largest-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn largest_number(nums: Vec) -> String { - let mut nums = nums - .into_iter() - .map(|num| num.to_string()) - .collect::>(); - nums.sort_unstable_by(|a, b| format!("{}{}", b, a).cmp(&format!("{}{}", a, b))); - if nums[0] == "0" { - return "0".to_owned(); - } - nums.iter().fold(String::new(), |mut s, num| { - s.push_str(num); - s - }) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_179() { - assert_eq!( - Solution::largest_number(vec![3, 30, 34, 5, 9]), - "9534330".to_owned() - ); - assert_eq!(Solution::largest_number(vec![121, 12]), "12121".to_owned()); - } -} diff --git a/src/solution/s0187_repeated_dna_sequences.rs b/src/solution/s0187_repeated_dna_sequences.rs deleted file mode 100644 index 9ceb86cb..00000000 --- a/src/solution/s0187_repeated_dna_sequences.rs +++ /dev/null @@ -1,100 +0,0 @@ -/** - * [187] Repeated DNA Sequences - * - * All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. - * - * Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule. - * - * Example: - * - * - * Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT" - * - * Output: ["AAAAACCCCC", "CCCCCAAAAA"] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/repeated-dna-sequences/ -// discuss: https://leetcode.com/problems/repeated-dna-sequences/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -首先想到直接长度为 10 的 sliding window 滑过去加一个 HashSet - -但这种方法在空间上和每次操作的耗时上都比较差, 可以转化为四进制或者二进制编码来考虑 - -A,C,G,T <-> [00, 01, 10, 11] - -那就简单很多了, 往后滑动一格不再需要调整整个 substring, 只需要移位, HashSet 也就存个 u32 即可 -*/ -use std::collections::HashSet; -impl Solution { - pub fn find_repeated_dna_sequences(s: String) -> Vec { - let mut seq_code: u32 = 0; - let mut set: HashSet = HashSet::new(); - let mut repeat: HashSet = HashSet::new(); - let mut count = 0; - for ch in s.chars() { - seq_code <<= 2; - match ch { - 'A' => seq_code |= 0_u32, - 'C' => seq_code |= 1_u32, - 'G' => seq_code |= 2_u32, - 'T' => seq_code |= 3_u32, - _ => unreachable!(), - } - // skip first 9 chars - if count < 9 { - count += 1; - continue; - } - // mask high 12-bits - seq_code &= 0b0000_0000_0000_1111_1111_1111_1111_1111; - if !set.insert(seq_code) { - repeat.insert(seq_code); - } - } - // bits code to seq string - repeat - .iter() - .map(|&code| { - let mut substr = String::new(); - let mut code = code; - for _ in 0..10 { - // take the first 2 bits each time - substr.push(match code & 0b0000_0000_0000_1100_0000_0000_0000_0000 { - 0b0000_0000_0000_0000_0000_0000_0000_0000 => 'A', - 0b0000_0000_0000_0100_0000_0000_0000_0000 => 'C', - 0b0000_0000_0000_1000_0000_0000_0000_0000 => 'G', - 0b0000_0000_0000_1100_0000_0000_0000_0000 => 'T', - _ => unreachable!(), - }); - code <<= 2; - } - substr - }) - .collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_187() { - assert_eq!( - Solution::find_repeated_dna_sequences("AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT".to_owned()), - vec_string!["AAAAACCCCC", "CCCCCAAAAA"] - ); - assert_eq!( - Solution::find_repeated_dna_sequences("GAGAGAGAGAGA".to_owned()), - vec_string!["GAGAGAGAGA"] - ); - } -} diff --git a/src/solution/s0188_best_time_to_buy_and_sell_stock_iv.rs b/src/solution/s0188_best_time_to_buy_and_sell_stock_iv.rs deleted file mode 100644 index e9c23319..00000000 --- a/src/solution/s0188_best_time_to_buy_and_sell_stock_iv.rs +++ /dev/null @@ -1,103 +0,0 @@ -/** - * [188] Best Time to Buy and Sell Stock IV - * - * Say you have an array for which the i^th element is the price of a given stock on day i. - * - * Design an algorithm to find the maximum profit. You may complete at most k transactions. - * - * Note:
- * You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). - * - * Example 1: - * - * - * Input: [2,4,1], k = 2 - * Output: 2 - * Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2. - * - * - * Example 2: - * - * - * Input: [3,2,6,5,0,3], k = 2 - * Output: 7 - * Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. - * Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/ -// discuss: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -已经在 #123 里解过了, 为了方便阅读直接把那题的分析拷贝到这里 - -先考虑只进行 1 次交易的情况, 我们求以 i *为售出点*, 只进行 1 次交易获得的最大利润, 那么: - -f[i] = if f[i-1] > 0 { f[i-1] } else { 0 } + prices[i] - prices[i-1] - -这很容易解, 解完之后找出 f 里的最大值即可, 但这不容易推广到 K 次交易的情况, 因为这时 f[i] 不代表到 i *为止*的最大利润, 无法作为单独的交易帮助递推 -(到 i 为止的含义是售出点可以在 [0,i] 之间) - -我们可以稍作改进, 变成求以 i 为结束点, 只进行 1 次交易获得的最大利润, 那么: - -f[i] = max( - f[i-1], - prices[i] - min(prices[j] { j in [0, i-1] }) -) - -这仍然是一个 O(N) 的解法, 因为 min(prices[j] { j in [0, i-1] }) 不需要遍历, 可以在递推过程中直接维护好 - -现在再推广到进行 K 次交易的情况, 那我们要求以 i 为结束点, 进行 k 次交易获得的最大利润, 这时有了变化, 我们可以在 j 之前再进行 K - 1 次交易: - -f[k, i] = max( - f[k, i-1], - prices[i] + max(f[k-1, j] - prices[j]) { j in [0, i-1] } ) -) - -显然, f[0, i] = 0, f[k, 0] = 0 - -这个算法可以形象地描述一下, 在 k = 1 时, 我们每次要找的就是 i 之前的最低谷点作为这次交易的开始点 j, 而当 k > 1 时, -我们 i 之前就有可能已经进行过交易了, 这时我们在找开始点 j 时, 就要同时考虑 "直到 j 为止, k-1 次交易的最大收益" - "j 本身的值". 以此来找到一个最佳点 j - -在实现时, 假如用 Bottom-Up 递推, 那么只需要维护一个 vec[i], 因为每轮递推时只会考虑上一轮的数据, 我们可以复用这个 O(N) 的额外存储空间 - -最后, 这题会给 k 非常大的 corner case, 实际上 k 大于 prices.len() / 2 后面就没有意义了, 可以 shortcut 掉(== 允许无穷次交易的场景), 下面写的比较糙, -直接限制了一下循环次数, 实际跑的时候运行时间会长一点 -*/ -impl Solution { - pub fn max_profit(k: i32, prices: Vec) -> i32 { - if prices.is_empty() { - return 0; - } - let max_trans = k as usize; - let mut cache = vec![0; prices.len()]; - for _ in 0..usize::min(max_trans, prices.len() / 2 + 1) { - // best_by_in 维护了考虑前 N 次交易的最佳的买入点, 即 max(f[k-1, j] - prices[j]) { j in [0, i-1] } - let mut best_buy_in = cache[0] - prices[0]; - for i in 1..prices.len() { - // 复用 vec 前暂存一下前一次的计算结果 - let temp = cache[i]; - cache[i] = i32::max(cache[i - 1], best_buy_in + prices[i]); - // 更新 best_buy_in - best_buy_in = i32::max(best_buy_in, temp - prices[i]); - } - } - return *cache.last().unwrap(); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_188() { - assert_eq!(Solution::max_profit(2, vec![3, 2, 6, 5, 0, 3]), 7); - } -} diff --git a/src/solution/s0189_rotate_array.rs b/src/solution/s0189_rotate_array.rs deleted file mode 100644 index 1a4d2ab8..00000000 --- a/src/solution/s0189_rotate_array.rs +++ /dev/null @@ -1,79 +0,0 @@ -/** - * [189] Rotate Array - * - * Given an array, rotate the array to the right by k steps, where k is non-negative. - * - * Example 1: - * - * - * Input: [1,2,3,4,5,6,7] and k = 3 - * Output: [5,6,7,1,2,3,4] - * Explanation: - * rotate 1 steps to the right: [7,1,2,3,4,5,6] - * rotate 2 steps to the right: [6,7,1,2,3,4,5] - * rotate 3 steps to the right: [5,6,7,1,2,3,4] - * - * - * Example 2: - * - * - * Input: [-1,-100,3,99] and k = 2 - * Output: [3,99,-1,-100] - * Explanation: - * rotate 1 steps to the right: [99,-1,-100,3] - * rotate 2 steps to the right: [3,99,-1,-100] - * - * - * Note: - * - * - * Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem. - * Could you do it in-place with O(1) extra space? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/rotate-array/ -// discuss: https://leetcode.com/problems/rotate-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn rotate(nums: &mut Vec, k: i32) { - let offset = (k as usize) % nums.len(); - if offset == 0 { - return; - } - let mut idx = 0; - let mut num = nums[0]; - let mut start_idx = 0; - for _ in 0..nums.len() { - idx = (idx + offset) % nums.len(); - let temp = num; - num = nums[idx]; - nums[idx] = temp; - if idx == start_idx { - idx += 1; - start_idx = idx; - num = nums[idx]; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_189() { - let mut nums = vec![1, 2, 3, 4, 5, 6, 7]; - Solution::rotate(&mut nums, 3); - assert_eq!(nums, vec![5, 6, 7, 1, 2, 3, 4]); - let mut nums = vec![1, 2, 3, 4, 5, 6]; - Solution::rotate(&mut nums, 2); - assert_eq!(nums, vec![5, 6, 1, 2, 3, 4]); - } -} diff --git a/src/solution/s0198_house_robber.rs b/src/solution/s0198_house_robber.rs deleted file mode 100644 index 446e63f1..00000000 --- a/src/solution/s0198_house_robber.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [198] House Robber - * - * You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night. - * - * Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police. - * - * Example 1: - * - * - * Input: [1,2,3,1] - * Output: 4 - * Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). - * Total amount you can rob = 1 + 3 = 4. - * - * Example 2: - * - * - * Input: [2,7,9,3,1] - * Output: 12 - * Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). - * Total amount you can rob = 2 + 9 + 1 = 12. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/house-robber/ -// discuss: https://leetcode.com/problems/house-robber/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -动态规划, 记抢到第 i 户为止的最大收益为 F[i], 则: - -i 有两种情况, 抢或不抢, 抢的话则最大收益是 F[i-2] + nums[i], -不抢则保持和前一次结束的收益一致, 等于 F[i-1], 于是: - -F[i] = i32::max(nums[i] + F[i-2], F[i-1]) - -观察到 F[i] 只依赖 F[i-1] 和 F[i-2], 可以用常数空间复杂度完成 -*/ -impl Solution { - pub fn rob(nums: Vec) -> i32 { - let mut former_max = 0; - let mut curr_max = 0; - for &num in nums.iter() { - let mut temp = curr_max; - curr_max = i32::max(former_max + num, curr_max); - former_max = temp; - } - curr_max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_198() { - assert_eq!(Solution::rob(vec![2, 7, 9, 3, 1]), 12); - assert_eq!(Solution::rob(vec![2, 7, 9, 10, 1]), 17); - assert_eq!(Solution::rob(vec![2, 1, 1, 2]), 4); - } -} diff --git a/src/solution/s0199_binary_tree_right_side_view.rs b/src/solution/s0199_binary_tree_right_side_view.rs deleted file mode 100644 index e07dc998..00000000 --- a/src/solution/s0199_binary_tree_right_side_view.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [199] Binary Tree Right Side View - * - * Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. - * - * Example: - * - * - * Input: [1,2,3,null,5,null,4] - * Output: [1, 3, 4] - * Explanation: - * - * 1 <--- - * / \ - * 2 3 <--- - * \ \ - * 5 4 <--- - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-right-side-view/ -// discuss: https://leetcode.com/problems/binary-tree-right-side-view/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::collections::VecDeque; -use std::rc::Rc; -impl Solution { - pub fn right_side_view(root: Option>>) -> Vec { - let mut res = Vec::new(); - let mut current_level = 0; - if root.is_none() { - return res; - } - let mut deq = VecDeque::new(); - deq.push_back((0, root.clone())); - res.push(root.as_ref().unwrap().borrow().val); - while !deq.is_empty() { - if let Some((level, Some(node))) = deq.pop_front() { - deq.push_back((level + 1, node.borrow().right.clone())); - deq.push_back((level + 1, node.borrow().left.clone())); - if level > current_level { - res.push(node.borrow().val); - current_level = level; - } - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_199() { - assert_eq!( - Solution::right_side_view(tree![1, 2, 3, null, 5, null, 4]), - vec![1, 3, 4] - ); - } -} diff --git a/src/solution/s0200_number_of_islands.rs b/src/solution/s0200_number_of_islands.rs deleted file mode 100644 index 262f0315..00000000 --- a/src/solution/s0200_number_of_islands.rs +++ /dev/null @@ -1,122 +0,0 @@ -/** - * [200] Number of Islands - * - * Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water. - * - * Example 1: - * - * - * Input: - * 11110 - * 11010 - * 11000 - * 00000 - * - * Output: 1 - * - * - * Example 2: - * - * - * Input: - * 11000 - * 11000 - * 00100 - * 00011 - * - * Output: 3 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/number-of-islands/ -// discuss: https://leetcode.com/problems/number-of-islands/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Union-Find Set -impl Solution { - pub fn num_islands(grid: Vec>) -> i32 { - if grid.is_empty() || grid[0].is_empty() { - return 0; - } - let (height, width) = (grid.len(), grid[0].len()); - let mut parent = vec![vec![(width, height); width]; height]; - for i in 0..height { - for j in 0..width { - if grid[i][j] != '1' { - continue; - } - parent[i][j] = (i, j); - if i > 0 && grid[i - 1][j] == '1' { - Solution::union(&mut parent, (i, j), (i - 1, j)); - } - if j > 0 && grid[i][j - 1] == '1' { - Solution::union(&mut parent, (i, j), (i, j - 1)); - } - } - } - let mut cnt = 0; - for i in 0..height { - for j in 0..width { - if parent[i][j] == (i, j) { - cnt += 1; - } - } - } - cnt - } - - fn get_parent(parent: &mut Vec>, p: (usize, usize)) -> (usize, usize) { - let mut child = p; - let mut p = p; - while parent[p.0][p.1] != p { - p = parent[p.0][p.1]; - } - // path compression, adjust all the node's parent to root along the path - while child != p { - let temp = parent[child.0][child.1]; - parent[child.0][child.1] = p; - child = temp; - } - p - } - - fn union(parent: &mut Vec>, p1: (usize, usize), p2: (usize, usize)) { - let p1 = Solution::get_parent(parent, p1); - let p2 = Solution::get_parent(parent, p2); - if p1 == p2 { - return; - } - parent[p1.0][p1.1] = p2 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_200() { - assert_eq!( - Solution::num_islands(vec![ - vec!['1', '1', '1', '1', '0',], - vec!['1', '1', '0', '1', '0',], - vec!['1', '1', '0', '0', '0',], - vec!['0', '0', '0', '0', '0',], - ]), - 1 - ); - assert_eq!( - Solution::num_islands(vec![ - vec!['1', '1', 'o', '1', '0',], - vec!['1', '1', '0', '1', '0',], - vec!['1', '1', '0', '0', '0',], - vec!['0', '0', '0', '1', '1',], - ]), - 3 - ); - } -} diff --git a/src/solution/s0201_bitwise_and_of_numbers_range.rs b/src/solution/s0201_bitwise_and_of_numbers_range.rs deleted file mode 100644 index 6c46e74f..00000000 --- a/src/solution/s0201_bitwise_and_of_numbers_range.rs +++ /dev/null @@ -1,58 +0,0 @@ -/** - * [201] Bitwise AND of Numbers Range - * - * Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive. - * - * Example 1: - * - * - * Input: [5,7] - * Output: 4 - * - * - * Example 2: - * - * - * Input: [0,1] - * Output: 0 - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/bitwise-and-of-numbers-range/ -// discuss: https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// just find the highest bit 1 of m and n -impl Solution { - pub fn range_bitwise_and(m: i32, n: i32) -> i32 { - let mut m = m; - let mut n = n; - if m == 0 { - return 0; - } - let mut step = 1; - while m != n { - // shortcut - if m == 0 { - return 0; - } - m >>= 1; - n >>= 1; - step <<= 1; - } - return m * step; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_201() { - assert_eq!(Solution::range_bitwise_and(5, 7), 4); - } -} diff --git a/src/solution/s0202_happy_number.rs b/src/solution/s0202_happy_number.rs deleted file mode 100644 index 076f2736..00000000 --- a/src/solution/s0202_happy_number.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [202] Happy Number - * - * Write an algorithm to determine if a number is "happy". - * - * A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers. - * - * Example: - * - * - * Input: 19 - * Output: true - * Explanation: - * 1^2 + 9^2 = 82 - * 8^2 + 2^2 = 68 - * 6^2 + 8^2 = 100 - * 1^2 + 0^2 + 0^2 = 1 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/happy-number/ -// discuss: https://leetcode.com/problems/happy-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashSet; -impl Solution { - pub fn is_happy(n: i32) -> bool { - let mut set = HashSet::new(); - let mut n = n; - loop { - set.insert(n); - let temp = Solution::next(n); - if temp == 1 { - return true; - } - if !set.insert(temp) { - return false; - } - n = temp - } - return false; - } - - fn next(n: i32) -> i32 { - let mut res = 0; - let mut n = n; - while n != 0 { - res += (n % 10).pow(2); - n = n / 10; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_202() { - assert_eq!(Solution::is_happy(19), true); - assert_eq!(Solution::is_happy(235123), false); - } -} diff --git a/src/solution/s0203_remove_linked_list_elements.rs b/src/solution/s0203_remove_linked_list_elements.rs deleted file mode 100644 index 3e48d811..00000000 --- a/src/solution/s0203_remove_linked_list_elements.rs +++ /dev/null @@ -1,50 +0,0 @@ -/** - * [203] Remove Linked List Elements - * - * Remove all elements from a linked list of integers that have value val. - * - * Example: - * - * - * Input: 1->2->6->3->4->5->6, val = 6 - * Output: 1->2->3->4->5 - * - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/remove-linked-list-elements/ -// discuss: https://leetcode.com/problems/remove-linked-list-elements/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn remove_elements(mut head: Option>, val: i32) -> Option> { - let mut dummy = Some(Box::new(ListNode::new(0))); - let mut next = dummy.as_mut(); - while let Some(mut inner) = head { - head = inner.next.take(); - if inner.val != val { - next.as_mut().unwrap().next = Some(inner); - next = next.unwrap().next.as_mut(); - } - } - dummy.unwrap().next - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_203() { - assert_eq!( - Solution::remove_elements(linked![1, 2, 6, 3, 4, 5, 6], 6), - linked![1, 2, 3, 4, 5] - ); - } -} diff --git a/src/solution/s0204_count_primes.rs b/src/solution/s0204_count_primes.rs deleted file mode 100644 index d6f0f134..00000000 --- a/src/solution/s0204_count_primes.rs +++ /dev/null @@ -1,68 +0,0 @@ -/** - * [204] Count Primes - * - * Count the number of prime numbers less than a non-negative number, n. - * - * Example: - * - * - * Input: 10 - * Output: 4 - * Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/count-primes/ -// discuss: https://leetcode.com/problems/count-primes/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn count_primes(n: i32) -> i32 { - if n <= 2 { - return 0; - } - let mut is_prime = vec![true; n as usize]; - is_prime[0] = false; - is_prime[1] = false; - let mut i = 2; - while i * i < n { - if !is_prime[i as usize] { - i += 1; - continue; - } - let mut j = i * i; - while j < n { - is_prime[j as usize] = false; - j += i; - } - i += 1; - } - let mut count = 0; - for &v in is_prime.iter() { - if v { - count += 1 - } - } - count - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_204() { - assert_eq!(Solution::count_primes(10), 4); - assert_eq!(Solution::count_primes(2), 0); - assert_eq!(Solution::count_primes(3), 1); - assert_eq!(Solution::count_primes(5), 2); - assert_eq!(Solution::count_primes(1), 0); - assert_eq!(Solution::count_primes(120), 30); - } -} diff --git a/src/solution/s0205_isomorphic_strings.rs b/src/solution/s0205_isomorphic_strings.rs deleted file mode 100644 index df697ebb..00000000 --- a/src/solution/s0205_isomorphic_strings.rs +++ /dev/null @@ -1,84 +0,0 @@ -/** - * [205] Isomorphic Strings - * - * Given two strings s and t, determine if they are isomorphic. - * - * Two strings are isomorphic if the characters in s can be replaced to get t. - * - * All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself. - * - * Example 1: - * - * - * Input: s = "egg", t = "add" - * Output: true - * - * - * Example 2: - * - * - * Input: s = "foo", t = "bar" - * Output: false - * - * Example 3: - * - * - * Input: s = "paper", t = "title" - * Output: true - * - * Note:
- * You may assume both s and t have the same length. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/isomorphic-strings/ -// discuss: https://leetcode.com/problems/isomorphic-strings/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::char; -use std::collections::hash_map::Entry; -use std::collections::HashMap; -impl Solution { - pub fn is_isomorphic(s: String, t: String) -> bool { - Solution::code(s) == Solution::code(t) - } - - fn code(s: String) -> String { - let mut map = HashMap::new(); - let mut start: char = '0'; - let mut res = String::new(); - for ch in s.chars() { - let v = map.entry(ch).or_insert_with(|| { - start = ((start as u8) + 1) as char; - start - }); - res.push(*v) - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_205() { - assert_eq!( - Solution::is_isomorphic("egg".to_owned(), "app".to_owned()), - true - ); - assert_eq!( - Solution::is_isomorphic("pecil".to_owned(), "this".to_owned()), - false - ); - assert_eq!( - Solution::is_isomorphic("paper".to_owned(), "title".to_owned()), - true - ); - } -} diff --git a/src/solution/s0206_reverse_linked_list.rs b/src/solution/s0206_reverse_linked_list.rs deleted file mode 100644 index b62aced8..00000000 --- a/src/solution/s0206_reverse_linked_list.rs +++ /dev/null @@ -1,52 +0,0 @@ -/** - * [206] Reverse Linked List - * - * Reverse a singly linked list. - * - * Example: - * - * - * Input: 1->2->3->4->5->NULL - * Output: 5->4->3->2->1->NULL - * - * - * Follow up: - * - * A linked list can be reversed either iteratively or recursively. Could you implement both? - * - */ -pub struct Solution {} -use crate::util::linked_list::{to_list, ListNode}; - -// problem: https://leetcode.com/problems/reverse-linked-list/ -// discuss: https://leetcode.com/problems/reverse-linked-list/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn reverse_list(head: Option>) -> Option> { - let mut curr = head; - let mut next = None; - while let Some(mut inner) = curr { - curr = inner.next.take(); - inner.next = next; - next = Some(inner); - } - next - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_206() { - assert_eq!( - Solution::reverse_list(linked![1, 2, 3, 4, 5]), - linked![5, 4, 3, 2, 1] - ); - } -} diff --git a/src/solution/s0207_course_schedule.rs b/src/solution/s0207_course_schedule.rs deleted file mode 100644 index 2a793683..00000000 --- a/src/solution/s0207_course_schedule.rs +++ /dev/null @@ -1,105 +0,0 @@ -/** - * [207] Course Schedule - * - * There are a total of n courses you have to take, labeled from 0 to n-1. - * - * Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] - * - * Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses? - * - * Example 1: - * - * - * Input: 2, [[1,0]] - * Output: true - * Explanation: There are a total of 2 courses to take. - * To take course 1 you should have finished course 0. So it is possible. - * - * Example 2: - * - * - * Input: 2, [[1,0],[0,1]] - * Output: false - * Explanation: There are a total of 2 courses to take. - * To take course 1 you should have finished course 0, and to take course 0 you should - * also have finished course 1. So it is impossible. - * - * - * Note: - * - *
    - * The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented. - * You may assume that there are no duplicate edges in the input prerequisites. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/course-schedule/ -// discuss: https://leetcode.com/problems/course-schedule/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// topology sort, BFS -use std::collections::VecDeque; -impl Solution { - pub fn can_finish(num_courses: i32, prerequisites: Vec>) -> bool { - let num = num_courses as usize; - let mut matrix = vec![vec![false; num]; num]; - let mut in_degree = vec![0; num]; - // collects node in degree - for pre in prerequisites.iter() { - if !matrix[pre[1] as usize][pre[0] as usize] { - in_degree[pre[0] as usize] += 1; - } - matrix[pre[1] as usize][pre[0] as usize] = true; - } - let mut deq = VecDeque::new(); - // BFS starts with nodes with 0 in degree - for (node, &v) in in_degree.iter().enumerate() { - if v == 0 { - deq.push_back(node); - } - } - let mut cnt = 0; - while let Some(node) = deq.pop_front() { - cnt += 1; - for (i, &connect) in matrix[node].iter().enumerate() { - if connect { - in_degree[i] -= 1; - if in_degree[i] == 0 { - deq.push_back(i); - } - } - } - } - cnt == num_courses - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_207() { - assert_eq!(Solution::can_finish(2, vec![vec![1, 0]]), true); - assert_eq!(Solution::can_finish(2, vec![vec![1, 0], vec![0, 1]]), false); - assert_eq!( - Solution::can_finish( - 8, - vec![ - vec![1, 0], - vec![2, 6], - vec![1, 7], - vec![6, 4], - vec![7, 0], - vec![0, 5] - ] - ), - true - ); - } -} diff --git a/src/solution/s0208_implement_trie_prefix_tree.rs b/src/solution/s0208_implement_trie_prefix_tree.rs deleted file mode 100644 index 9ac8e470..00000000 --- a/src/solution/s0208_implement_trie_prefix_tree.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [208] Implement Trie (Prefix Tree) - * - * Implement a trie with insert, search, and startsWith methods. - * - * Example: - * - * - * Trie trie = new Trie(); - * - * trie.insert("apple"); - * trie.search("apple"); // returns true - * trie.search("app"); // returns false - * trie.startsWith("app"); // returns true - * trie.insert("app"); - * trie.search("app"); // returns true - * - * - * Note: - * - * - * You may assume that all inputs are consist of lowercase letters a-z. - * All inputs are guaranteed to be non-empty strings. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/implement-trie-prefix-tree/ -// discuss: https://leetcode.com/problems/implement-trie-prefix-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -#[derive(Default)] -struct Trie { - is_end: bool, - nodes: [Option>; 26], -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl Trie { - /** Initialize your data structure here. */ - fn new() -> Self { - Default::default() - } - - /** insert a word into the trie. */ - fn insert(&mut self, word: String) { - let mut curr = self; - - for i in word.chars().map(|ch| (ch as u8 - 'a' as u8) as usize) { - curr = curr.nodes[i].get_or_insert_with(|| Box::new(Trie::new())); - } - curr.is_end = true; - } - - /** Returns if the word is in the trie. */ - fn search(&self, word: String) -> bool { - self.find(word).map_or(false, |t| t.is_end) - } - - /** Returns if there is any word in the trie that starts with the given prefix. */ - fn starts_with(&self, prefix: String) -> bool { - self.find(prefix).is_some() - } - - fn find(&self, word: String) -> Option<&Trie> { - let mut curr = self; - for i in word.chars().map(|ch| (ch as u8 - 'a' as u8) as usize) { - curr = curr.nodes[i].as_ref()?; - } - Some(curr) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_208() { - let mut trie = Trie::new(); - trie.insert("apple".to_owned()); - assert_eq!(trie.search("apple".to_owned()), true); // returns true - assert_eq!(trie.search("app".to_owned()), false); - assert_eq!(trie.starts_with("app".to_owned()), true); // returns true - trie.insert("app".to_owned()); - assert_eq!(trie.search("app".to_owned()), true); // returns true - } -} diff --git a/src/solution/s0209_minimum_size_subarray_sum.rs b/src/solution/s0209_minimum_size_subarray_sum.rs deleted file mode 100644 index b6a692d7..00000000 --- a/src/solution/s0209_minimum_size_subarray_sum.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [209] Minimum Size Subarray Sum - * - * Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead. - * - * Example: - * - * - * Input: s = 7, nums = [2,3,1,2,4,3] - * Output: 2 - * Explanation: the subarray [4,3] has the minimal length under the problem constraint. - * - *
Follow up:
- * - *
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/minimum-size-subarray-sum/ -// discuss: https://leetcode.com/problems/minimum-size-subarray-sum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn min_sub_array_len(s: i32, nums: Vec) -> i32 { - let (mut i, mut j) = (0_usize, 0_usize); - let mut min = i32::max_value(); - let mut found = false; - let mut sum = 0; - while j < nums.len() { - sum += nums[j]; - if sum >= s { - found = true; - while i <= j { - sum -= nums[i]; - i += 1; - if sum < s { - min = i32::min(min, j as i32 - i as i32 + 2); - break; - } - } - } - j += 1; - } - if found { - min - } else { - 0 - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_209() { - assert_eq!(Solution::min_sub_array_len(7, vec![2, 3, 1, 2, 4, 3]), 2); - assert_eq!(Solution::min_sub_array_len(4, vec![1, 4, 4]), 1); - } -} diff --git a/src/solution/s0210_course_schedule_ii.rs b/src/solution/s0210_course_schedule_ii.rs deleted file mode 100644 index dd3a1b03..00000000 --- a/src/solution/s0210_course_schedule_ii.rs +++ /dev/null @@ -1,98 +0,0 @@ -/** - * [210] Course Schedule II - * - * There are a total of n courses you have to take, labeled from 0 to n-1. - * - * Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] - * - * Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses. - * - * There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array. - * - * Example 1: - * - * - * Input: 2, [[1,0]] - * Output: [0,1] - * Explanation: There are a total of 2 courses to take. To take course 1 you should have finished - * course 0. So the correct course order is [0,1] . - * - * Example 2: - * - * - * Input: 4, [[1,0],[2,0],[3,1],[3,2]] - * Output: [0,1,2,3] or [0,2,1,3] - * Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both - * courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. - * So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] . - * - * Note: - * - *
    - * The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented. - * You may assume that there are no duplicate edges in the input prerequisites. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/course-schedule-ii/ -// discuss: https://leetcode.com/problems/course-schedule-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::VecDeque; -impl Solution { - pub fn find_order(num_courses: i32, prerequisites: Vec>) -> Vec { - let num = num_courses as usize; - let mut matrix = vec![vec![false; num]; num]; - let mut in_degree = vec![0; num]; - // collects node in degree - for pre in prerequisites.iter() { - if !matrix[pre[1] as usize][pre[0] as usize] { - in_degree[pre[0] as usize] += 1; - } - matrix[pre[1] as usize][pre[0] as usize] = true; - } - let mut deq = VecDeque::new(); - // BFS starts with nodes with 0 in degree - for (node, &v) in in_degree.iter().enumerate() { - if v == 0 { - deq.push_back(node); - } - } - let mut res = Vec::with_capacity(num); - while let Some(node) = deq.pop_front() { - res.push(node as i32); - for (i, &connect) in matrix[node].iter().enumerate() { - if connect { - in_degree[i] -= 1; - if in_degree[i] == 0 { - deq.push_back(i); - } - } - } - } - if res.len() == num { - res - } else { - vec![] - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_210() { - assert_eq!(Solution::find_order(2, vec![vec![1, 0]]), vec![0, 1]); - assert_eq!( - Solution::find_order(4, vec![vec![1, 0], vec![2, 0], vec![3, 1], vec![3, 2]]), - vec![0, 1, 2, 3] - ); - } -} diff --git a/src/solution/s0211_add_and_search_word_data_structure_design.rs b/src/solution/s0211_add_and_search_word_data_structure_design.rs deleted file mode 100644 index 0652f951..00000000 --- a/src/solution/s0211_add_and_search_word_data_structure_design.rs +++ /dev/null @@ -1,129 +0,0 @@ -/** - * [211] Add and Search Word - Data structure design - * - * Design a data structure that supports the following two operations: - * - * - * void addWord(word) - * bool search(word) - * - * - * search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter. - * - * Example: - * - * - * addWord("bad") - * addWord("dad") - * addWord("mad") - * search("pad") -> false - * search("bad") -> true - * search(".ad") -> true - * search("b..") -> true - * - * - * Note:
- * You may assume that all words are consist of lowercase letters a-z. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/add-and-search-word-data-structure-design/ -// discuss: https://leetcode.com/problems/add-and-search-word-data-structure-design/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct WordDictionary { - root: Option>, -} - -#[derive(Default)] -struct Trie { - is_end: bool, - marked: Vec, - nodes: [Option>; 26], -} - -impl Trie { - fn new() -> Self { - Default::default() - } -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl WordDictionary { - // /** Initialize your data structure here. */ - // fn new() -> Self { - // WordDictionary{ - // root: Some(Box::new(Trie::new())), - // } - // } - - // /** Adds a word into the data structure. */ - // fn add_word(&mut self, word: String) { - // let mut curr = self.root; - - // for i in word.chars().map(|ch| (ch as u8 - 'a' as u8) as usize) { - // curr = curr.as_ref().unwrap().nodes[i]; - // } - // curr.as_mut().unwrap().is_end = true; - // } - - // /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */ - // fn search(&self, word: String) -> bool { - // let mut chs: Vec = word.chars().collect(); - // WordDictionary::search_from(self.root.as_ref(), &mut chs) - // } - - // fn search_from(node: Option<&Box>, chs: &mut [char]) -> bool { - // if node.is_none() { - // return false - // } - // let node = node.unwrap(); - // if chs.len() < 1 { - // return node.is_end - // } - // if chs[0] == '.' { - // // backtracking - // let mut sliced = &mut chs[1..]; - // for &idx in node.marked.iter() { - // if WordDictionary::search_from(node.nodes[idx].as_ref(), sliced) { - // return true - // } - // } - // false - // } else { - // let next = node.nodes[(chs[0] as u8 - 'a' as u8) as usize].as_ref(); - // WordDictionary::search_from(next, &mut chs[1..]) - // } - // } -} - -/** - * Your WordDictionary object will be instantiated and called as such: - * let obj = WordDictionary::new(); - * obj.add_word(word); - * let ret_2: bool = obj.search(word); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_211() { - // let mut dict = WordDictionary::new(); - // dict.add_word("bad".to_owned()); - // dict.add_word("dad".to_owned()); - // dict.add_word("mad".to_owned()); - // assert_eq!(dict.search("pad".to_owned()), false); - // assert_eq!(dict.search("bad".to_owned()), true); - // assert_eq!(dict.search(".ad".to_owned()), true); - // assert_eq!(dict.search("da.".to_owned()), true); - } -} diff --git a/src/solution/s0212_word_search_ii.rs b/src/solution/s0212_word_search_ii.rs deleted file mode 100644 index ada034c8..00000000 --- a/src/solution/s0212_word_search_ii.rs +++ /dev/null @@ -1,57 +0,0 @@ -/** - * [212] Word Search II - * - * Given a 2D board and a list of words from the dictionary, find all words in the board. - * - * Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word. - * - * - * - * Example: - * - * - * Input: - * board = [ - * ['o','a','a','n'], - * ['e','t','a','e'], - * ['i','h','k','r'], - * ['i','f','l','v'] - * ] - * words = ["oath","pea","eat","rain"] - * - * Output: ["eat","oath"] - * - * - * - * - * Note: - * - *
    - * All inputs are consist of lowercase letters a-z. - * The values of words are distinct. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-search-ii/ -// discuss: https://leetcode.com/problems/word-search-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -// TODO - -impl Solution { - pub fn find_words(board: Vec>, words: Vec) -> Vec { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_212() {} -} diff --git a/src/solution/s0213_house_robber_ii.rs b/src/solution/s0213_house_robber_ii.rs deleted file mode 100644 index fbe55040..00000000 --- a/src/solution/s0213_house_robber_ii.rs +++ /dev/null @@ -1,69 +0,0 @@ -/** - * [213] House Robber II - * - * You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night. - * - * Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police. - * - * Example 1: - * - * - * Input: [2,3,2] - * Output: 3 - * Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), - * because they are adjacent houses. - * - * - * Example 2: - * - * - * Input: [1,2,3,1] - * Output: 4 - * Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). - * Total amount you can rob = 1 + 3 = 4. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/house-robber-ii/ -// discuss: https://leetcode.com/problems/house-robber-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// DP twice: rob first one || not rob first one -// F[i] = max(F[i-2] + num[i], F[i-1]) -impl Solution { - pub fn rob(nums: Vec) -> i32 { - let mut max = i32::min_value(); - for &rob_first in vec![true, false].iter() { - let (mut prev, mut curr) = (0, 0); - for (k, &num) in nums.iter().enumerate() { - if k == 0 && !rob_first { - continue; - } - // k is last element but not the first element - if k != 0 && k == (nums.len() - 1) && rob_first { - continue; - } - let next = i32::max(prev + num, curr); - prev = curr; - curr = next; - } - max = i32::max(max, curr) - } - max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_213() { - assert_eq!(Solution::rob(vec![2, 3, 2]), 3); - assert_eq!(Solution::rob(vec![1, 2, 3, 1]), 4); - } -} diff --git a/src/solution/s0214_shortest_palindrome.rs b/src/solution/s0214_shortest_palindrome.rs deleted file mode 100644 index 425eec95..00000000 --- a/src/solution/s0214_shortest_palindrome.rs +++ /dev/null @@ -1,40 +0,0 @@ -/** - * [214] Shortest Palindrome - * - * Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. - * - * Example 1: - * - * - * Input: "aacecaaa" - * Output: "aaacecaaa" - * - * - * Example 2: - * - * - * Input: "abcd" - * Output: "dcbabcd" - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/shortest-palindrome/ -// discuss: https://leetcode.com/problems/shortest-palindrome/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn shortest_palindrome(s: String) -> String { - "".to_owned() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_214() {} -} diff --git a/src/solution/s0215_kth_largest_element_in_an_array.rs b/src/solution/s0215_kth_largest_element_in_an_array.rs deleted file mode 100644 index 159e7af6..00000000 --- a/src/solution/s0215_kth_largest_element_in_an_array.rs +++ /dev/null @@ -1,62 +0,0 @@ -/** - * [215] Kth Largest Element in an Array - * - * Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element. - * - * Example 1: - * - * - * Input: [3,2,1,5,6,4] and k = 2 - * Output: 5 - * - * - * Example 2: - * - * - * Input: [3,2,3,1,2,4,5,5,6] and k = 4 - * Output: 4 - * - * Note:
- * You may assume k is always valid, 1 ≤ k ≤ array's length. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/kth-largest-element-in-an-array/ -// discuss: https://leetcode.com/problems/kth-largest-element-in-an-array/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cmp::Reverse; -use std::collections::BinaryHeap; -impl Solution { - pub fn find_kth_largest(nums: Vec, k: i32) -> i32 { - let k = k as usize; - let mut heap = BinaryHeap::with_capacity(k); - for &num in nums.iter() { - if heap.len() < k as usize { - heap.push(Reverse(num)) - } else if num >= heap.peek().unwrap().0 { - heap.pop(); - heap.push(Reverse(num)); - } - } - heap.peek().unwrap().0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_215() { - assert_eq!( - Solution::find_kth_largest(vec![3, 2, 3, 1, 2, 4, 5, 5, 6], 4), - 4 - ); - assert_eq!(Solution::find_kth_largest(vec![3, 2, 1, 5, 6, 4], 2), 5); - } -} diff --git a/src/solution/s0216_combination_sum_iii.rs b/src/solution/s0216_combination_sum_iii.rs deleted file mode 100644 index 0f714e01..00000000 --- a/src/solution/s0216_combination_sum_iii.rs +++ /dev/null @@ -1,83 +0,0 @@ -/** - * [216] Combination Sum III - * - *
- * Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. - * - * Note: - * - * - * All numbers will be positive integers. - * The solution set must not contain duplicate combinations. - * - * - * Example 1: - * - * - * Input: k = 3, n = 7 - * Output: [[1,2,4]] - * - * - * Example 2: - * - * - * Input: k = 3, n = 9 - * Output: [[1,2,6], [1,3,5], [2,3,4]] - * - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/combination-sum-iii/ -// discuss: https://leetcode.com/problems/combination-sum-iii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn combination_sum3(k: i32, n: i32) -> Vec> { - if k > 9 || k < 1 { - return vec![]; - } - let max = (0..k).fold(0, |acc, t| acc + 9 - t); - let min = (0..k).fold(0, |acc, t| acc + t); - if n < min || n > max { - return vec![]; - } - let mut res = Vec::new(); - let mut seed = Vec::new(); - Solution::helper(n, 0, k, seed, &mut res); - res - } - - fn helper(distance: i32, prev: i32, remain: i32, mut curr: Vec, res: &mut Vec>) { - if remain == 0 { - if distance == 0 { - res.push(curr); - } - return; - } - for i in (prev + 1..=9) { - if distance - i < 0 { - break; - } - let mut new_vec = curr.clone(); - new_vec.push(i); - Solution::helper(distance - i, i, remain - 1, new_vec, res); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_216() { - assert_eq!( - Solution::combination_sum3(3, 9), - vec![vec![1, 2, 6], vec![1, 3, 5], vec![2, 3, 4]] - ); - } -} diff --git a/src/solution/s0217_contains_duplicate.rs b/src/solution/s0217_contains_duplicate.rs deleted file mode 100644 index 0d2e5e12..00000000 --- a/src/solution/s0217_contains_duplicate.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [217] Contains Duplicate - * - * Given an array of integers, find if the array contains any duplicates. - * - * Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. - * - * Example 1: - * - * - * Input: [1,2,3,1] - * Output: true - * - * Example 2: - * - * - * Input: [1,2,3,4] - * Output: false - * - * Example 3: - * - * - * Input: [1,1,1,3,3,4,3,2,4,2] - * Output: true - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/contains-duplicate/ -// discuss: https://leetcode.com/problems/contains-duplicate/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn contains_duplicate(nums: Vec) -> bool { - if nums.is_empty() { - return false; - } - let mut nums = nums; - nums.sort_unstable(); - let mut prev = nums[0]; - for i in 1..nums.len() { - if nums[i] == prev { - return true; - } - prev = nums[i] - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_217() { - assert_eq!(Solution::contains_duplicate(vec![1]), false); - assert_eq!(Solution::contains_duplicate(vec![]), false); - assert_eq!(Solution::contains_duplicate(vec![1, 2, 3, 4]), false); - assert_eq!(Solution::contains_duplicate(vec![1, 2, 3, 1]), true); - } -} diff --git a/src/solution/s0218_the_skyline_problem.rs b/src/solution/s0218_the_skyline_problem.rs deleted file mode 100644 index 7f0ba3c1..00000000 --- a/src/solution/s0218_the_skyline_problem.rs +++ /dev/null @@ -1,46 +0,0 @@ -/** - * [218] The Skyline Problem - * - * A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are given the locations and height of all the buildings as shown on a cityscape photo (Figure A), write a program to output the skyline formed by these buildings collectively (Figure B). - * Buildings Skyline Contour - * - * The geometric information of each building is represented by a triplet of integers [Li, Ri, Hi], where Li and Ri are the x coordinates of the left and right edge of the ith building, respectively, and Hi is its height. It is guaranteed that 0 ≤ Li, Ri ≤ INT_MAX, 0 < Hi ≤ INT_MAX, and Ri - Li > 0. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0. - * - * For instance, the dimensions of all buildings in Figure A are recorded as: [ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ] . - * - * The output is a list of "key points" (red dots in Figure B) in the format of [ [x1,y1], [x2, y2], [x3, y3], ... ] that uniquely defines a skyline. A key point is the left endpoint of a horizontal line segment. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour. - * - * For instance, the skyline in Figure B should be represented as:[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ]. - * - * Notes: - * - * - * The number of buildings in any input list is guaranteed to be in the range [0, 10000]. - * The input list is already sorted in ascending order by the left x position Li. - * The output list must be sorted by the x position. - * There must be no consecutive horizontal lines of equal height in the output skyline. For instance, [...[2 3], [4 5], [7 5], [11 5], [12 7]...] is not acceptable; the three lines of height 5 should be merged into one in the final output as such: [...[2 3], [4 5], [12 7], ...] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/the-skyline-problem/ -// discuss: https://leetcode.com/problems/the-skyline-problem/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn get_skyline(buildings: Vec>) -> Vec> { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_218() {} -} diff --git a/src/solution/s0219_contains_duplicate_ii.rs b/src/solution/s0219_contains_duplicate_ii.rs deleted file mode 100644 index da446dae..00000000 --- a/src/solution/s0219_contains_duplicate_ii.rs +++ /dev/null @@ -1,79 +0,0 @@ -/** - * [219] Contains Duplicate II - * - * Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k. - * - *
- * Example 1: - * - * - * Input: nums = [1,2,3,1], k = 3 - * Output: true - * - * - *
- * Example 2: - * - * - * Input: nums = [1,0,1,1], k = 1 - * Output: true - * - * - *
- * Example 3: - * - * - * Input: nums = [1,2,3,1,2,3], k = 2 - * Output: false - * - *
- *
- *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/contains-duplicate-ii/ -// discuss: https://leetcode.com/problems/contains-duplicate-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashMap; -impl Solution { - pub fn contains_nearby_duplicate(nums: Vec, k: i32) -> bool { - let mut map = HashMap::new(); - for (idx, &num) in nums.iter().enumerate() { - match map.get(&num) { - Some(v) => { - if idx - v <= k as usize { - return true; - } - map.insert(num, idx); - } - None => { - map.insert(num, idx); - } - } - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_219() { - assert_eq!( - Solution::contains_nearby_duplicate(vec![1, 2, 3, 1, 2, 3], 2), - false - ); - assert_eq!( - Solution::contains_nearby_duplicate(vec![1, 2, 3, 1, 2, 3], 3), - true - ); - } -} diff --git a/src/solution/s0220_contains_duplicate_iii.rs b/src/solution/s0220_contains_duplicate_iii.rs deleted file mode 100644 index 0677b1d3..00000000 --- a/src/solution/s0220_contains_duplicate_iii.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [220] Contains Duplicate III - * - * Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k. - * - *
- * Example 1: - * - * - * Input: nums = [1,2,3,1], k = 3, t = 0 - * Output: true - * - * - *
- * Example 2: - * - * - * Input: nums = [1,0,1,1], k = 1, t = 2 - * Output: true - * - * - *
- * Example 3: - * - * - * Input: nums = [1,5,9,1,5,9], k = 2, t = 3 - * Output: false - * - *
- *
- *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/contains-duplicate-iii/ -// discuss: https://leetcode.com/problems/contains-duplicate-iii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -use std::collections::HashMap; -impl Solution { - pub fn contains_nearby_almost_duplicate(nums: Vec, k: i32, t: i32) -> bool { - if k < 1 || t < 0 { - return false; - } - let mut map = HashMap::new(); - for i in 0..nums.len() { - let remap = nums[i] as i64 - i32::min_value() as i64; - let bucket = remap / (t as i64 + 1); - println!("{} {}", remap, bucket); - if map.contains_key(&bucket) - || map - .get(&(bucket - 1)) - .map_or(false, |v| remap - v <= t as i64) - || map - .get(&(bucket + 1)) - .map_or(false, |v| v - remap <= t as i64) - { - return true; - } - if i >= k as usize { - let last_bucket = - (nums[i - k as usize] as i64 - i32::min_value() as i64) / (t as i64 + 1); - map.remove(&last_bucket); - } - map.insert(bucket, remap); - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_220() { - // assert_eq!(Solution::contains_nearby_almost_duplicate(vec![1,5,9,1,5,9], 2, 3), false); - // assert_eq!(Solution::contains_nearby_almost_duplicate(vec![1,2,3,1], 3, 0), true); - assert_eq!( - Solution::contains_nearby_almost_duplicate(vec![-1, 2147483647], 1, 2147483647), - false - ); - } -} diff --git a/src/solution/s0221_maximal_square.rs b/src/solution/s0221_maximal_square.rs deleted file mode 100644 index 34f793ac..00000000 --- a/src/solution/s0221_maximal_square.rs +++ /dev/null @@ -1,84 +0,0 @@ -/** - * [221] Maximal Square - * - * Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area. - * - * Example: - * - * - * Input: - * - * 1 0 1 0 0 - * 1 0 1 1 1 - * 1 1 1 1 1 - * 1 0 0 1 0 - * - * Output: 4 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/maximal-square/ -// discuss: https://leetcode.com/problems/maximal-square/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -DP, f(i, j) to represent the max square of matrix that end with (i, j) (right bottom corener), then: - -f(0, 0) = matrix[0][0] -f(i, j) = if matrix[0][0] { min(f(i-1,j), f(i,j-1), f(i-1)(j-1)) + 1 } else { 0 } - -The equation explained: - -matrix: dp: -1 1 1 1 1 1 -1 1 1 -> 1 2 2 -1 1 1 1 2 3 -*/ -impl Solution { - pub fn maximal_square(matrix: Vec>) -> i32 { - if matrix.is_empty() || matrix[0].is_empty() { - return 0; - } - let (height, width) = (matrix.len(), matrix[0].len()); - let mut dp = vec![vec![0; width]; height]; - let mut max = 0; - for i in 0..height { - for j in 0..width { - if matrix[i][j] == '0' { - continue; - } - dp[i][j] = i32::min( - i32::min( - if i < 1 { 0 } else { dp[i - 1][j] }, - if j < 1 { 0 } else { dp[i][j - 1] }, - ), - if i < 1 || j < 1 { 0 } else { dp[i - 1][j - 1] }, - ) + 1; - max = i32::max(max, dp[i][j]) - } - } - max * max - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_221() { - assert_eq!( - Solution::maximal_square(vec![ - vec!['1', '0', '1', '0', '0'], - vec!['1', '0', '1', '1', '1'], - vec!['1', '1', '1', '1', '1'], - vec!['1', '0', '0', '1', '0'], - ]), - 4 - ) - } -} diff --git a/src/solution/s0222_count_complete_tree_nodes.rs b/src/solution/s0222_count_complete_tree_nodes.rs deleted file mode 100644 index ae0d410b..00000000 --- a/src/solution/s0222_count_complete_tree_nodes.rs +++ /dev/null @@ -1,88 +0,0 @@ -/** - * [222] Count Complete Tree Nodes - * - * Given a complete binary tree, count the number of nodes. - * - * Note: - * - * Definition of a complete binary tree from Wikipedia:
- * In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2^h nodes inclusive at the last level h. - * - * Example: - * - * - * Input: - * 1 - * / \ - * 2 3 - * / \ / - * 4 5 6 - * - * Output: 6 - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/count-complete-tree-nodes/ -// discuss: https://leetcode.com/problems/count-complete-tree-nodes/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn count_nodes(root: Option>>) -> i32 { - // 0. get the hight of full nodes - let mut height = 0; - let mut curr = root.clone(); - let mut result = 0; - while let Some(inner) = curr { - height += 1; - result += 2_i32.pow(height - 1); - curr = inner.borrow().right.clone(); - } - if height == 0 { - return result; - } - // 1. 'binary search' to find the node number of the lowest level - // the lowest level may have 0~2^H-1 node - let mut curr_root = root.clone(); - while height > 1 { - // see if left tree is full - let mut node = curr_root.clone().unwrap().borrow().left.clone(); - let mut level = 2; - while level < height { - node = node.unwrap().borrow().right.clone(); - level += 1; - } - if node.unwrap().borrow().right.is_some() { - curr_root = curr_root.unwrap().borrow().right.clone(); - result += 2_i32.pow(height - 1); - } else { - curr_root = curr_root.unwrap().borrow().left.clone(); - } - height -= 1; - } - if curr_root.as_ref().unwrap().borrow().left.is_some() { - result += 1; - } - result - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_222() { - assert_eq!(Solution::count_nodes(tree![1, 1, 1, 1, 1, 1, 1]), 7); - assert_eq!(Solution::count_nodes(tree![]), 0); - assert_eq!(Solution::count_nodes(tree![1, 1]), 2); - assert_eq!(Solution::count_nodes(tree![1]), 1); - assert_eq!(Solution::count_nodes(tree![1, 1, 1]), 3); - } -} diff --git a/src/solution/s0223_rectangle_area.rs b/src/solution/s0223_rectangle_area.rs deleted file mode 100644 index 647b6b2a..00000000 --- a/src/solution/s0223_rectangle_area.rs +++ /dev/null @@ -1,62 +0,0 @@ -/** - * [223] Rectangle Area - * - * Find the total area covered by two rectilinear rectangles in a 2D plane. - * - * Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. - * - * Rectangle Area - * - * Example: - * - * - * Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2 - * Output: 45 - * - * Note: - * - * Assume that the total area is never beyond the maximum possible value of int. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/rectangle-area/ -// discuss: https://leetcode.com/problems/rectangle-area/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// mention the integer divition -impl Solution { - pub fn compute_area(a: i32, b: i32, c: i32, d: i32, e: i32, f: i32, g: i32, h: i32) -> i32 { - let center1 = ((a + c), (b + d)); - let center2 = ((e + g), (f + h)); - let rect1 = (c - a, d - b); - let rect2 = (g - e, h - f); - let x_intersect = i32::min( - (rect1.0 + rect2.0 - (center1.0 - center2.0).abs()), - i32::min(2 * rect1.0, 2 * rect2.0), - ) / 2; - let x_intersect = if x_intersect < 0 { 0 } else { x_intersect }; - let y_intersect = i32::min( - (rect1.1 + rect2.1 - (center1.1 - center2.1).abs()), - i32::min(2 * rect1.1, 2 * rect2.1), - ) / 2; - let y_intersect = if y_intersect < 0 { 0 } else { y_intersect }; - rect1.0 * rect1.1 - x_intersect * y_intersect + rect2.0 * rect2.1 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_223() { - assert_eq!(Solution::compute_area(0, 0, 0, 0, 0, 0, 0, 0), 0); - assert_eq!(Solution::compute_area(-3, 0, 3, 4, 0, -1, 9, 2), 45); - assert_eq!(Solution::compute_area(-2, -2, 2, 2, -2, -2, 2, 2), 16); - assert_eq!(Solution::compute_area(-2, -2, 2, 2, -1, 4, 1, 6), 20); - } -} diff --git a/src/solution/s0224_basic_calculator.rs b/src/solution/s0224_basic_calculator.rs deleted file mode 100644 index 5d917661..00000000 --- a/src/solution/s0224_basic_calculator.rs +++ /dev/null @@ -1,156 +0,0 @@ -/** - * [224] Basic Calculator - * - * Implement a basic calculator to evaluate a simple expression string. - * - * The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . - * - * Example 1: - * - * - * Input: "1 + 1" - * Output: 2 - * - * - * Example 2: - * - * - * Input: " 2-1 + 2 " - * Output: 3 - * - * Example 3: - * - * - * Input: "(1+(4+5+2)-3)+(6+8)" - * Output: 23 - * Note: - * - * - * You may assume that the given expression is always valid. - * Do not use the eval built-in library function. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/basic-calculator/ -// discuss: https://leetcode.com/problems/basic-calculator/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -#[derive(PartialEq, Copy, Clone, Debug)] -enum Token { - LeftBracket, - RightBracket, - PlusSign, - MinusSign, - Number(i64), -} - -impl Solution { - pub fn calculate(s: String) -> i32 { - // lexer - let mut token_stream = Vec::new(); - let mut num = 0_i64; - let mut in_num = false; - for ch in s.chars() { - match ch { - '0'..='9' => { - in_num = true; - num = 10 * num + (ch as u8 - '0' as u8) as i64; - } - _ => { - if in_num { - token_stream.push(Token::Number(num)); - num = 0; - in_num = false; - } - match ch { - '(' => { - token_stream.push(Token::LeftBracket); - } - ')' => { - token_stream.push(Token::RightBracket); - } - '+' => { - token_stream.push(Token::PlusSign); - } - '-' => { - token_stream.push(Token::MinusSign); - } - _ => {} - }; - } - } - } - if in_num { - token_stream.push(Token::Number(num)); - } - - // parser - let mut stack = Vec::new(); - let mut iter = token_stream.into_iter(); - let mut pause = false; - let mut token = Token::LeftBracket; - loop { - if !pause { - token = if let Some(token) = iter.next() { - token - } else { - break; - } - } else { - pause = false; - } - match token { - Token::LeftBracket => { - stack.push(token); - } - Token::RightBracket => { - if let Token::Number(right_hand) = stack.pop().unwrap() { - stack.pop(); - pause = true; - token = Token::Number(right_hand); - } - } - Token::PlusSign => { - stack.push(token); - } - Token::MinusSign => { - stack.push(token); - } - Token::Number(num) => { - if stack.is_empty() || Token::LeftBracket == *stack.last().unwrap() { - stack.push(Token::Number(num)); - } else { - let sign = stack.pop().unwrap(); - if let Token::Number(left_hand) = stack.pop().unwrap() { - let res = - left_hand + num * if Token::PlusSign == sign { 1 } else { -1 }; - stack.push(Token::Number(res)); - } - } - } - } - } - if let Token::Number(num) = stack.pop().unwrap() { - return num as i32; - } - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_224() { - assert_eq!(Solution::calculate("(1+(4+5+2)-3)+(6+8)".to_owned()), 23); - assert_eq!(Solution::calculate("1+1".to_owned()), 2); - assert_eq!(Solution::calculate("0".to_owned()), 0); - assert_eq!(Solution::calculate("2147483647".to_owned()), 2147483647); - } -} diff --git a/src/solution/s0225_implement_stack_using_queues.rs b/src/solution/s0225_implement_stack_using_queues.rs deleted file mode 100644 index f162e9a6..00000000 --- a/src/solution/s0225_implement_stack_using_queues.rs +++ /dev/null @@ -1,125 +0,0 @@ -/** - * [225] Implement Stack using Queues - * - * Implement the following operations of a stack using queues. - * - * - * push(x) -- Push element x onto stack. - * pop() -- Removes the element on top of the stack. - * top() -- Get the top element. - * empty() -- Return whether the stack is empty. - * - * - * Example: - * - * - * MyStack stack = new MyStack(); - * - * stack.push(1); - * stack.push(2); - * stack.top(); // returns 2 - * stack.pop(); // returns 2 - * stack.empty(); // returns false - * - * Notes: - * - * - * You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid. - * Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue. - * You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack). - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/implement-stack-using-queues/ -// discuss: https://leetcode.com/problems/implement-stack-using-queues/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct MyStack { - q1: VecDeque, - q2: VecDeque, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -use std::collections::VecDeque; -impl MyStack { - /** Initialize your data structure here. */ - fn new() -> Self { - MyStack { - q1: VecDeque::new(), - q2: VecDeque::new(), - } - } - - /** Push element x onto stack. */ - fn push(&mut self, x: i32) { - if self.q1.is_empty() { - self.q1.push_back(x); - while let Some(v) = self.q2.pop_front() { - self.q1.push_back(v); - } - } else { - self.q2.push_back(x); - while let Some(v) = self.q1.pop_front() { - self.q2.push_back(v); - } - } - } - - /** Removes the element on top of the stack and returns that element. */ - fn pop(&mut self) -> i32 { - if self.q1.is_empty() { - self.q2.pop_front().unwrap() - } else { - self.q1.pop_front().unwrap() - } - } - - /** Get the top element. */ - fn top(&self) -> i32 { - if self.q1.is_empty() { - *self.q2.front().unwrap() - } else { - *self.q1.front().unwrap() - } - } - - /** Returns whether the stack is empty. */ - fn empty(&self) -> bool { - self.q1.is_empty() && self.q2.is_empty() - } -} - -/** - * Your MyStack object will be instantiated and called as such: - * let obj = MyStack::new(); - * obj.push(x); - * let ret_2: i32 = obj.pop(); - * let ret_3: i32 = obj.top(); - * let ret_4: bool = obj.empty(); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_225() { - let mut stack = MyStack::new(); - - stack.push(1); - stack.push(2); - assert_eq!(stack.top(), 2); // returns 2 - assert_eq!(stack.pop(), 2); // returns 2 - assert_eq!(stack.empty(), false); // returns false - assert_eq!(stack.pop(), 1); - assert_eq!(stack.empty(), true); - } -} diff --git a/src/solution/s0226_invert_binary_tree.rs b/src/solution/s0226_invert_binary_tree.rs deleted file mode 100644 index 1a1b6115..00000000 --- a/src/solution/s0226_invert_binary_tree.rs +++ /dev/null @@ -1,69 +0,0 @@ -/** - * [226] Invert Binary Tree - * - * Invert a binary tree. - * - * Example: - * - * Input: - * - * - * 4 - * / \ - * 2 7 - * / \ / \ - * 1 3 6 9 - * - * Output: - * - * - * 4 - * / \ - * 7 2 - * / \ / \ - * 9 6 3 1 - * - * Trivia:
- * This problem was inspired by this original tweet by Max Howell: - * - *
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.
- * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/invert-binary-tree/ -// discuss: https://leetcode.com/problems/invert-binary-tree/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn invert_tree(root: Option>>) -> Option>> { - if let Some(node) = root.clone() { - Solution::invert_tree(node.borrow().right.clone()); - Solution::invert_tree(node.borrow().left.clone()); - let left = node.borrow().left.clone(); - let right = node.borrow().right.clone(); - node.borrow_mut().left = right; - node.borrow_mut().right = left; - } - root - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_226() { - assert_eq!( - Solution::invert_tree(tree![4, 2, 7, 1, 3, 6, 9]), - tree![4, 7, 2, 9, 6, 3, 1] - ); - } -} diff --git a/src/solution/s0227_basic_calculator_ii.rs b/src/solution/s0227_basic_calculator_ii.rs deleted file mode 100644 index 07e349dd..00000000 --- a/src/solution/s0227_basic_calculator_ii.rs +++ /dev/null @@ -1,107 +0,0 @@ -/** - * [227] Basic Calculator II - * - * Implement a basic calculator to evaluate a simple expression string. - * - * The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. - * - * Example 1: - * - * - * Input: "3+2*2" - * Output: 7 - * - * - * Example 2: - * - * - * Input: " 3/2 " - * Output: 1 - * - * Example 3: - * - * - * Input: " 3+5 / 2 " - * Output: 5 - * - * - * Note: - * - * - * You may assume that the given expression is always valid. - * Do not use the eval built-in library function. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/basic-calculator-ii/ -// discuss: https://leetcode.com/problems/basic-calculator-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn calculate(s: String) -> i32 { - let mut acc = 0_i64; - let mut prev = 0_i64; - let mut curr = 0_i64; - let mut sign = 1; - let mut has_prev = false; - let mut multiple = true; - for ch in s.chars() { - match ch { - '0'..='9' => { - curr = 10 * curr + (ch as u8 - '0' as u8) as i64; - } - '+' | '-' => { - if has_prev { - if multiple { - curr = prev * curr; - } else { - curr = prev / curr; - } - has_prev = false; - } - acc += curr * sign; - curr = 0; - sign = if ch == '+' { 1 } else { -1 }; - } - '*' | '/' => { - if has_prev { - if multiple { - curr = prev * curr; - } else { - curr = prev / curr; - } - } - has_prev = true; - prev = curr; - curr = 0; - multiple = if ch == '*' { true } else { false }; - } - _ => {} - } - } - if has_prev { - if multiple { - curr = prev * curr; - } else { - curr = prev / curr; - } - } - acc += sign * curr; - acc as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_227() { - assert_eq!(Solution::calculate("3+2*2".to_owned()), 7); - } -} diff --git a/src/solution/s0228_summary_ranges.rs b/src/solution/s0228_summary_ranges.rs deleted file mode 100644 index 63e8fe77..00000000 --- a/src/solution/s0228_summary_ranges.rs +++ /dev/null @@ -1,74 +0,0 @@ -/** - * [228] Summary Ranges - * - * Given a sorted integer array without duplicates, return the summary of its ranges. - * - * Example 1: - * - * - * Input: [0,1,2,4,5,7] - * Output: ["0->2","4->5","7"] - * Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range. - * - * - * Example 2: - * - * - * Input: [0,2,3,4,6,8,9] - * Output: ["0","2->4","6","8->9"] - * Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/summary-ranges/ -// discuss: https://leetcode.com/problems/summary-ranges/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn summary_ranges(nums: Vec) -> Vec { - if nums.is_empty() { - return vec![]; - } - let mut res = Vec::new(); - let mut curr = nums[0]; - let mut start = nums[0]; - for num in nums.into_iter().skip(1) { - if num == curr + 1 { - curr = num; - } else { - // seq done - Solution::record(&mut res, start, curr); - start = num; - curr = num; - } - } - Solution::record(&mut res, start, curr); - res - } - - pub fn record(vec: &mut Vec, start: i32, end: i32) { - if start == end { - vec.push(format!("{}", start)) - } else { - vec.push(format!("{}->{}", start, end)) - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_228() { - assert_eq!( - Solution::summary_ranges(vec![0, 1, 2, 3, 4, 5, 6]), - vec_string!["0->6"] - ); - } -} diff --git a/src/solution/s0229_majority_element_ii.rs b/src/solution/s0229_majority_element_ii.rs deleted file mode 100644 index fab987ba..00000000 --- a/src/solution/s0229_majority_element_ii.rs +++ /dev/null @@ -1,87 +0,0 @@ -/** - * [229] Majority Element II - * - * Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. - * - * Note: The algorithm should run in linear time and in O(1) space. - * - * Example 1: - * - * - * Input: [3,2,3] - * Output: [3] - * - * Example 2: - * - * - * Input: [1,1,1,3,3,2,2,2] - * Output: [1,2] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/majority-element-ii/ -// discuss: https://leetcode.com/problems/majority-element-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn majority_element(nums: Vec) -> Vec { - if nums.is_empty() { - return vec![]; - } - let (mut vote0, mut vote1, mut candidate0, mut candidate1) = (0, 0, -1, -2); - for &num in nums.iter() { - if num == candidate0 { - vote0 += 1; - } else if num == candidate1 { - vote1 += 1; - } else if vote0 == 0 { - candidate0 = num; - vote0 = 1; - } else if vote1 == 0 { - candidate1 = num; - vote1 = 1; - } else { - vote0 -= 1; - vote1 -= 1; - } - } - // the presents of majority element is not guaranteed, we have to do a double check - let mut res = Vec::new(); - for &v in vec![candidate0, candidate1].iter() { - let mut count = 0; - for &num in nums.iter() { - if v == num { - count += 1; - } - } - if count > (nums.len() / 3) as i32 { - res.push(v); - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_229() { - assert_eq!( - Solution::majority_element(vec![1, 1, 1, 2, 2, 2, 3, 3, 3]), - vec![] - ); - assert_eq!( - Solution::majority_element(vec![1, 1, 1, 2, 2, 3, 3, 3]), - vec![1, 3] - ); - assert_eq!(Solution::majority_element(vec![1]), vec![1]); - assert_eq!(Solution::majority_element(vec![5, 6, 6]), vec![6]); - assert_eq!(Solution::majority_element(vec![1, 2, 3, 4]), vec![]); - } -} diff --git a/src/solution/s0230_kth_smallest_element_in_a_bst.rs b/src/solution/s0230_kth_smallest_element_in_a_bst.rs deleted file mode 100644 index 0533f705..00000000 --- a/src/solution/s0230_kth_smallest_element_in_a_bst.rs +++ /dev/null @@ -1,84 +0,0 @@ -/** - * [230] Kth Smallest Element in a BST - * - * Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. - * - * Note:
- * You may assume k is always valid, 1 ≤ k ≤ BST's total elements. - * - * Example 1: - * - * - * Input: root = [3,1,4,null,2], k = 1 - * 3 - * / \ - * 1 4 - * \ - * 2 - * Output: 1 - * - * Example 2: - * - * - * Input: root = [5,3,6,2,4,null,null,1], k = 3 - * 5 - * / \ - * 3 6 - * / \ - * 2 4 - * / - * 1 - * Output: 3 - * - * - * Follow up:
- * What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine? - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/kth-smallest-element-in-a-bst/ -// discuss: https://leetcode.com/problems/kth-smallest-element-in-a-bst/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn kth_smallest(root: Option>>, k: i32) -> i32 { - let mut res = 0; - Solution::helper(root, k, &mut res); - res - } - - pub fn helper(root: Option>>, k: i32, res: &mut i32) -> i32 { - if k <= 0 { - return 0; - } - if let Some(node) = root { - let left = Solution::helper(node.borrow().left.clone(), k, res); - if left == 1 { - *res = node.borrow().val; - } - let right = Solution::helper(node.borrow().right.clone(), left - 1, res); - right - } else { - k - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_230() { - assert_eq!(Solution::kth_smallest(tree![3, 1, 4, null, 2], 1), 1); - assert_eq!(Solution::kth_smallest(tree![3, 1, 4, null, 2], 2), 2); - assert_eq!(Solution::kth_smallest(tree![3, 1, 4, null, 2], 3), 3); - } -} diff --git a/src/solution/s0231_power_of_two.rs b/src/solution/s0231_power_of_two.rs deleted file mode 100644 index 334fc314..00000000 --- a/src/solution/s0231_power_of_two.rs +++ /dev/null @@ -1,54 +0,0 @@ -/** - * [231] Power of Two - * - * Given an integer, write a function to determine if it is a power of two. - * - * Example 1: - * - * - * Input: 1 - * Output: true - * Explanation: 2^0 = 1 - * - * - * Example 2: - * - * - * Input: 16 - * Output: true - * Explanation: 2^4 = 16 - * - * Example 3: - * - * - * Input: 218 - * Output: false - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/power-of-two/ -// discuss: https://leetcode.com/problems/power-of-two/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_power_of_two(n: i32) -> bool { - let results: Vec = (0..31).map(|x| 2_i32.pow(x)).collect(); - results.binary_search(&n).is_ok() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_231() { - assert_eq!(Solution::is_power_of_two(-1), false); - assert_eq!(Solution::is_power_of_two(1), true); - assert_eq!(Solution::is_power_of_two(1024), true); - } -} diff --git a/src/solution/s0232_implement_queue_using_stacks.rs b/src/solution/s0232_implement_queue_using_stacks.rs deleted file mode 100644 index 70bf3735..00000000 --- a/src/solution/s0232_implement_queue_using_stacks.rs +++ /dev/null @@ -1,110 +0,0 @@ -/** - * [232] Implement Queue using Stacks - * - * Implement the following operations of a queue using stacks. - * - * - * push(x) -- Push element x to the back of queue. - * pop() -- Removes the element from in front of queue. - * peek() -- Get the front element. - * empty() -- Return whether the queue is empty. - * - * - * Example: - * - * - * MyQueue queue = new MyQueue(); - * - * queue.push(1); - * queue.push(2); - * queue.peek(); // returns 1 - * queue.pop(); // returns 1 - * queue.empty(); // returns false - * - * Notes: - * - * - * You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid. - * Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack. - * You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue). - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/implement-queue-using-stacks/ -// discuss: https://leetcode.com/problems/implement-queue-using-stacks/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct MyQueue { - vec1: Vec, - vec2: Vec, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl MyQueue { - /** Initialize your data structure here. */ - fn new() -> Self { - MyQueue { - vec1: Vec::new(), - vec2: Vec::new(), - } - } - - /** Push element x to the back of queue. */ - fn push(&mut self, x: i32) { - while let Some(v) = self.vec1.pop() { - self.vec2.push(v); - } - self.vec2.push(x); - while let Some(v) = self.vec2.pop() { - self.vec1.push(v); - } - } - - /** Removes the element from in front of queue and returns that element. */ - fn pop(&mut self) -> i32 { - self.vec1.pop().unwrap() - } - - /** Get the front element. */ - fn peek(&self) -> i32 { - *self.vec1.last().unwrap() - } - - /** Returns whether the queue is empty. */ - fn empty(&self) -> bool { - self.vec1.is_empty() - } -} - -/** - * Your MyQueue object will be instantiated and called as such: - * let obj = MyQueue::new(); - * obj.push(x); - * let ret_2: i32 = obj.pop(); - * let ret_3: i32 = obj.peek(); - * let ret_4: bool = obj.empty(); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_232() { - let mut queue = MyQueue::new(); - - queue.push(1); - queue.push(2); - assert_eq!(queue.peek(), 1); // returns 1 - assert_eq!(queue.pop(), 1); // returns 1 - assert_eq!(queue.empty(), false); // returns false - } -} diff --git a/src/solution/s0233_number_of_digit_one.rs b/src/solution/s0233_number_of_digit_one.rs deleted file mode 100644 index 60097ed0..00000000 --- a/src/solution/s0233_number_of_digit_one.rs +++ /dev/null @@ -1,36 +0,0 @@ -/** - * [233] Number of Digit One - * - * Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. - * - * Example: - * - * - * Input: 13 - * Output: 6 - * Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/number-of-digit-one/ -// discuss: https://leetcode.com/problems/number-of-digit-one/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn count_digit_one(n: i32) -> i32 { - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_233() {} -} diff --git a/src/solution/s0238_product_of_array_except_self.rs b/src/solution/s0238_product_of_array_except_self.rs deleted file mode 100644 index ef5683c4..00000000 --- a/src/solution/s0238_product_of_array_except_self.rs +++ /dev/null @@ -1,65 +0,0 @@ -/** - * [238] Product of Array Except Self - * - * Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. - * - * Example: - * - * - * Input: [1,2,3,4] - * Output: [24,12,8,6] - * - * - * Note: Please solve it without division and in O(n). - * - * Follow up:
- * Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.) - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/product-of-array-except-self/ -// discuss: https://leetcode.com/problems/product-of-array-except-self/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -x 2 3 4 = 24 -1 x 3 4 = 12 -1 2 x 4 = 8 -1 2 3 x = 6 - */ -impl Solution { - pub fn product_except_self(nums: Vec) -> Vec { - if nums.len() < 2 { - return vec![]; - } - let mut res = vec![1; nums.len()]; - let mut n = 1; - for i in (0..nums.len() - 1).rev() { - n *= nums[i + 1]; - res[i] = n; - } - n = 1; - for i in 1..nums.len() { - n *= nums[i - 1]; - res[i] *= n; - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_238() { - assert_eq!( - Solution::product_except_self(vec![1, 2, 3, 4]), - vec![24, 12, 8, 6] - ); - } -} diff --git a/src/solution/s0239_sliding_window_maximum.rs b/src/solution/s0239_sliding_window_maximum.rs deleted file mode 100644 index f7509cb9..00000000 --- a/src/solution/s0239_sliding_window_maximum.rs +++ /dev/null @@ -1,95 +0,0 @@ -/** - * [239] Sliding Window Maximum - * - * Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window. - * - * Example: - * - * - * Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3 - * Output: [3,3,5,5,6,7] - * Explanation: - * - * Window position Max - * --------------- ----- - * [1 3 -1] -3 5 3 6 7 3 - * 1 [3 -1 -3] 5 3 6 7 3 - * 1 3 [-1 -3 5] 3 6 7 5 - * 1 3 -1 [-3 5 3] 6 7 5 - * 1 3 -1 -3 [5 3 6] 7 6 - * 1 3 -1 -3 5 [3 6 7] 7 - * - * - * Note:
- * You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array. - * - * Follow up:
- * Could you solve it in linear time? - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/sliding-window-maximum/ -// discuss: https://leetcode.com/problems/sliding-window-maximum/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::VecDeque; -impl Solution { - pub fn max_sliding_window(nums: Vec, k: i32) -> Vec { - let mut deq: VecDeque<(usize, i32)> = VecDeque::new(); - let mut res = Vec::new(); - for i in 0..nums.len() { - // maintain sliding window - if !deq.is_empty() && (*deq.front().unwrap()).0 as i32 <= (i as i32) - k { - deq.pop_front(); - let mut max = i32::min_value(); - let mut count = 0_usize; - for (j, &v) in deq.iter().enumerate() { - if v.1 >= max { - max = v.1; - count = j; - } - } - for _ in 0..count { - deq.pop_front(); - } - } - let num = nums[i]; - if !deq.is_empty() && (*deq.front().unwrap()).1 <= num { - while !deq.is_empty() { - deq.pop_front(); - } - } - deq.push_back((i, num)); - if i + 1 >= k as usize { - res.push((*deq.front().unwrap()).1); - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_239() { - assert_eq!( - Solution::max_sliding_window(vec![9, 10, 9, -7, -4, -8, 2, -6], 5), - vec![10, 10, 9, 2] - ); - assert_eq!( - Solution::max_sliding_window(vec![1, 3, 1, 2, 0, 5], 3), - vec![3, 3, 2, 5] - ); - assert_eq!(Solution::max_sliding_window(vec![7, 2, 4], 2), vec![7, 4]); - assert_eq!(Solution::max_sliding_window(vec![1, -1], 1), vec![1, -1]); - assert_eq!( - Solution::max_sliding_window(vec![1, 3, -1, -3, 5, 3, 6, 7], 3), - vec![3, 3, 5, 5, 6, 7] - ); - } -} diff --git a/src/solution/s0241_different_ways_to_add_parentheses.rs b/src/solution/s0241_different_ways_to_add_parentheses.rs deleted file mode 100644 index a76bfa6a..00000000 --- a/src/solution/s0241_different_ways_to_add_parentheses.rs +++ /dev/null @@ -1,81 +0,0 @@ -/** - * [241] Different Ways to Add Parentheses - * - * Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *. - * - * Example 1: - * - * - * Input: "2-1-1" - * Output: [0, 2] - * Explanation: - * ((2-1)-1) = 0 - * (2-(1-1)) = 2 - * - * Example 2: - * - * - * Input: "2*3-4*5" - * Output: [-34, -14, -10, -10, 10] - * Explanation: - * (2*(3-(4*5))) = -34 - * ((2*3)-(4*5)) = -14 - * ((2*(3-4))*5) = -10 - * (2*((3-4)*5)) = -10 - * (((2*3)-4)*5) = 10 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/different-ways-to-add-parentheses/ -// discuss: https://leetcode.com/problems/different-ways-to-add-parentheses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn diff_ways_to_compute(input: String) -> Vec { - Solution::helper(&input) - } - - pub fn helper(input: &str) -> Vec { - if input.is_empty() { - return vec![]; - } - if let Ok(digit) = input.parse::() { - return vec![digit]; - } - let mut res: Vec = Vec::new(); - for (i, ch) in input.chars().enumerate() { - if ch == '+' || ch == '-' || ch == '*' { - let left = Solution::helper(&input[..i]); - let right = Solution::helper(&input[i + 1..]); - for &a in left.iter() { - for &b in right.iter() { - res.push(match ch { - '+' => a + b, - '-' => a - b, - '*' => a * b, - _ => unreachable!(), - }) - } - } - } - } - res - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_241() { - assert_eq!( - Solution::diff_ways_to_compute("2*3-4*5".to_owned()), - vec![-34, -10, -14, -10, 10] - ); - } -} diff --git a/src/solution/s0242_valid_anagram.rs b/src/solution/s0242_valid_anagram.rs deleted file mode 100644 index ba7a58ae..00000000 --- a/src/solution/s0242_valid_anagram.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [242] Valid Anagram - * - * Given two strings s and t , write a function to determine if t is an anagram of s. - * - * Example 1: - * - * - * Input: s = "anagram", t = "nagaram" - * Output: true - * - * - * Example 2: - * - * - * Input: s = "rat", t = "car" - * Output: false - * - * - * Note:
- * You may assume the string contains only lowercase alphabets. - * - * Follow up:
- * What if the inputs contain unicode characters? How would you adapt your solution to such case? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/valid-anagram/ -// discuss: https://leetcode.com/problems/valid-anagram/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_anagram(s: String, t: String) -> bool { - Solution::hit(s) == Solution::hit(t) - } - - fn hit(s: String) -> Vec { - let mut hit = vec![0; 27]; - for ch in s.chars() { - hit[(ch as u8 - 'a' as u8) as usize] += 1; - } - hit - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_242() { - assert_eq!( - Solution::is_anagram("anagram".to_owned(), "nagaram".to_owned()), - true - ); - } -} diff --git a/src/solution/s0257_binary_tree_paths.rs b/src/solution/s0257_binary_tree_paths.rs deleted file mode 100644 index 6c7833e2..00000000 --- a/src/solution/s0257_binary_tree_paths.rs +++ /dev/null @@ -1,85 +0,0 @@ -/** - * [257] Binary Tree Paths - * - * Given a binary tree, return all root-to-leaf paths. - * - * Note: A leaf is a node with no children. - * - * Example: - * - * - * Input: - * - * 1 - * / \ - * 2 3 - * \ - * 5 - * - * Output: ["1->2->5", "1->3"] - * - * Explanation: All root-to-leaf paths are: 1->2->5, 1->3 - * - */ -pub struct Solution {} -use crate::util::tree::{to_tree, TreeNode}; - -// problem: https://leetcode.com/problems/binary-tree-paths/ -// discuss: https://leetcode.com/problems/binary-tree-paths/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// Definition for a binary tree node. -// #[derive(Debug, PartialEq, Eq)] -// pub struct TreeNode { -// pub val: i32, -// pub left: Option>>, -// pub right: Option>>, -// } -// -// impl TreeNode { -// #[inline] -// pub fn new(val: i32) -> Self { -// TreeNode { -// val, -// left: None, -// right: None -// } -// } -// } -use std::cell::RefCell; -use std::rc::Rc; -impl Solution { - pub fn binary_tree_paths(root: Option>>) -> Vec { - let mut res = Vec::new(); - Solution::helper(root, "".to_owned(), &mut res); - res - } - - fn helper(root: Option>>, path: String, res: &mut Vec) { - if let Some(inner) = root { - if inner.borrow().left.is_none() && inner.borrow().right.is_none() { - res.push(format!("{}{}", path, inner.borrow().val)); - } else { - let path = format!("{}{}->", path, inner.borrow().val); - Solution::helper(inner.borrow().left.clone(), path.clone(), res); - Solution::helper(inner.borrow().right.clone(), path, res); - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_257() { - assert_eq!( - Solution::binary_tree_paths(tree![1, 2, 3, null, 5]), - vec_string!["1->2->5", "1->3"] - ); - } -} diff --git a/src/solution/s0258_add_digits.rs b/src/solution/s0258_add_digits.rs deleted file mode 100644 index 2248be1d..00000000 --- a/src/solution/s0258_add_digits.rs +++ /dev/null @@ -1,41 +0,0 @@ -/** - * [258] Add Digits - * - * Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. - * - * Example: - * - * - * Input: 38 - * Output: 2 - * Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. - * Since 2 has only one digit, return it. - * - * - * Follow up:
- * Could you do it without any loop/recursion in O(1) runtime? - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/add-digits/ -// discuss: https://leetcode.com/problems/add-digits/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn add_digits(num: i32) -> i32 { - 1 + ((num - 1) % 9) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_258() { - assert_eq!(Solution::add_digits(1234), 1); - } -} diff --git a/src/solution/s0260_single_number_iii.rs b/src/solution/s0260_single_number_iii.rs deleted file mode 100644 index 09d7a920..00000000 --- a/src/solution/s0260_single_number_iii.rs +++ /dev/null @@ -1,57 +0,0 @@ -/** - * [260] Single Number III - * - * Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. - * - * Example: - * - * - * Input: [1,2,1,3,2,5] - * Output: [3,5] - * - * Note: - * - *
    - * The order of the result is not important. So in the above example, [5, 3] is also correct. - * Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity? - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/single-number-iii/ -// discuss: https://leetcode.com/problems/single-number-iii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn single_number(nums: Vec) -> Vec { - let mut res = 0; - for &num in nums.iter() { - res = res ^ num; - } - let right_most_set_bit = res & !(res - 1); - let mut bit_set = 0; - let mut bit_unset = 0; - for &num in nums.iter() { - if num & right_most_set_bit == 0 { - bit_unset = bit_unset ^ num; - } else { - bit_set = bit_set ^ num; - } - } - return vec![bit_set, bit_unset]; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_260() { - assert_eq!(Solution::single_number(vec![1, 2, 1, 2, 3, 4]), vec![3, 4]); - assert_eq!(Solution::single_number(vec![1, 2, 1, 3, 2, 5]), vec![3, 5]); - } -} diff --git a/src/solution/s0263_ugly_number.rs b/src/solution/s0263_ugly_number.rs deleted file mode 100644 index 9ef3bf57..00000000 --- a/src/solution/s0263_ugly_number.rs +++ /dev/null @@ -1,73 +0,0 @@ -/** - * [263] Ugly Number - * - * Write a program to check whether a given number is an ugly number. - * - * Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. - * - * Example 1: - * - * - * Input: 6 - * Output: true - * Explanation: 6 = 2 × 3 - * - * Example 2: - * - * - * Input: 8 - * Output: true - * Explanation: 8 = 2 × 2 × 2 - * - * - * Example 3: - * - * - * Input: 14 - * Output: false - * Explanation: 14 is not ugly since it includes another prime factor 7. - * - * - * Note: - * - *
    - * 1 is typically treated as an ugly number. - * Input is within the 32-bit signed integer range: [-2^31, 2^31 - 1]. - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/ugly-number/ -// discuss: https://leetcode.com/problems/ugly-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn is_ugly(num: i32) -> bool { - if num <= 0 { - false - } else if num == 1 { - true - } else if num % 5 == 0 { - Solution::is_ugly(num / 5) - } else if num % 3 == 0 { - Solution::is_ugly(num / 3) - } else if num % 2 == 0 { - Solution::is_ugly(num / 2) - } else { - false - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_263() { - assert_eq!(Solution::is_ugly(25), true); - } -} diff --git a/src/solution/s0264_ugly_number_ii.rs b/src/solution/s0264_ugly_number_ii.rs deleted file mode 100644 index 8dda4f01..00000000 --- a/src/solution/s0264_ugly_number_ii.rs +++ /dev/null @@ -1,66 +0,0 @@ -/** - * [264] Ugly Number II - * - * Write a program to find the n-th ugly number. - * - * Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. - * - * Example: - * - * - * Input: n = 10 - * Output: 12 - * Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers. - * - * Note: - * - *
    - * 1 is typically treated as an ugly number. - * n does not exceed 1690. - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/ugly-number-ii/ -// discuss: https://leetcode.com/problems/ugly-number-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn nth_ugly_number(n: i32) -> i32 { - let mut base2 = 0; - let mut base3 = 0; - let mut base5 = 0; - let mut ugly = vec![1; 1]; - for _ in 1..n { - let next2 = ugly[base2] * 2; - let next3 = ugly[base3] * 3; - let next5 = ugly[base5] * 5; - let next = i32::min(next2, i32::min(next3, next5)); - if next2 == next { - base2 += 1 - } - if next3 == next { - base3 += 1 - } - if next5 == next { - base5 += 1 - } - ugly.push(next) - } - *ugly.last().unwrap() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_264() { - assert_eq!(Solution::nth_ugly_number(1), 1); - assert_eq!(Solution::nth_ugly_number(10), 12); - } -} diff --git a/src/solution/s0268_missing_number.rs b/src/solution/s0268_missing_number.rs deleted file mode 100644 index 63291ee0..00000000 --- a/src/solution/s0268_missing_number.rs +++ /dev/null @@ -1,46 +0,0 @@ -/** - * [268] Missing Number - * - * Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. - * - * Example 1: - * - * - * Input: [3,0,1] - * Output: 2 - * - * - * Example 2: - * - * - * Input: [9,6,4,2,3,5,7,0,1] - * Output: 8 - * - * - * Note:
- * Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity? - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/missing-number/ -// discuss: https://leetcode.com/problems/missing-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn missing_number(nums: Vec) -> i32 { - ((nums.len() + 1) * nums.len()) as i32 / 2 - nums.into_iter().fold(0, |acc, v| acc + v) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_268() { - assert_eq!(Solution::missing_number(vec![3, 0, 1]), 2); - } -} diff --git a/src/solution/s0273_integer_to_english_words.rs b/src/solution/s0273_integer_to_english_words.rs deleted file mode 100644 index 123a1f9d..00000000 --- a/src/solution/s0273_integer_to_english_words.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [273] Integer to English Words - * - * Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 2^31 - 1. - * - * Example 1: - * - * - * Input: 123 - * Output: "One Hundred Twenty Three" - * - * - * Example 2: - * - * - * Input: 12345 - * Output: "Twelve Thousand Three Hundred Forty Five" - * - * Example 3: - * - * - * Input: 1234567 - * Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven" - * - * - * Example 4: - * - * - * Input: 1234567891 - * Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One" - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/integer-to-english-words/ -// discuss: https://leetcode.com/problems/integer-to-english-words/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// static digits: Vec<&str> = vec!["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"]; -// static xteens: Vec<&str> = vec!["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventenn", "Eighteen", "Nineteen"]; -// static xtys: Vec<&str> = vec!["Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]; -// static hunderd: &str = "Hunderd"; -// static thousands: Vec<&str> = vec!["Thousand", "Million", "Billion"]; - -impl Solution { - pub fn number_to_words(num: i32) -> String { - "".to_owned() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_273() {} -} diff --git a/src/solution/s0274_h_index.rs b/src/solution/s0274_h_index.rs deleted file mode 100644 index b36f65a3..00000000 --- a/src/solution/s0274_h_index.rs +++ /dev/null @@ -1,53 +0,0 @@ -/** - * [274] H-Index - * - * Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index. - * - * According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N - h papers have no more than h citations each." - * - * Example: - * - * - * Input: citations = [3,0,6,1,5] - * Output: 3 - * Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had - * received 3, 0, 6, 1, 5 citations respectively. - * Since the researcher has 3 papers with at least 3 citations each and the remaining - * two with no more than 3 citations each, her h-index is 3. - * - * Note: If there are several possible values for h, the maximum one is taken as the h-index. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/h-index/ -// discuss: https://leetcode.com/problems/h-index/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn h_index(mut citations: Vec) -> i32 { - citations.sort_unstable(); - let mut hidx = 0; - for &v in citations.iter().rev() { - if v >= hidx + 1 { - hidx += 1; - } else { - break; - } - } - hidx - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_274() { - assert_eq!(Solution::h_index(vec![3, 0, 6, 1, 5]), 3); - } -} diff --git a/src/solution/s0275_h_index_ii.rs b/src/solution/s0275_h_index_ii.rs deleted file mode 100644 index 027ad02a..00000000 --- a/src/solution/s0275_h_index_ii.rs +++ /dev/null @@ -1,74 +0,0 @@ -/** - * [275] H-Index II - * - * Given an array of citations sorted in ascending order (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index. - * - * According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N - h papers have no more than h citations each." - * - * Example: - * - * - * Input: citations = [0,1,3,5,6] - * Output: 3 - * Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had - * received 0, 1, 3, 5, 6 citations respectively. - * Since the researcher has 3 papers with at least 3 citations each and the remaining - * two with no more than 3 citations each, her h-index is 3. - * - * Note: - * - * If there are several possible values for h, the maximum one is taken as the h-index. - * - * Follow up: - * - * - * This is a follow up problem to H-Index, where citations is now guaranteed to be sorted in ascending order. - * Could you solve it in logarithmic time complexity? - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/h-index-ii/ -// discuss: https://leetcode.com/problems/h-index-ii/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn h_index(citations: Vec) -> i32 { - if citations.is_empty() { - return 0; - } - let len = citations.len(); - let (mut low, mut high) = (0_usize, len - 1); - while low <= high { - let mid = low + (high - low) / 2; - if citations[mid] >= (len - mid) as i32 { - if mid == 0 { - break; - } - high = mid - 1; - } else { - low = mid + 1; - } - } - (len - low) as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_275() { - assert_eq!(Solution::h_index(vec![]), 0); - assert_eq!(Solution::h_index(vec![0]), 0); - assert_eq!(Solution::h_index(vec![11, 15]), 2); - assert_eq!(Solution::h_index(vec![1]), 1); - assert_eq!(Solution::h_index(vec![0, 1, 3, 5, 6]), 3); - assert_eq!(Solution::h_index(vec![0, 4, 4, 5, 6]), 4); - } -} diff --git a/src/solution/s0279_perfect_squares.rs b/src/solution/s0279_perfect_squares.rs deleted file mode 100644 index fe5f0a01..00000000 --- a/src/solution/s0279_perfect_squares.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [279] Perfect Squares - * - * Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. - * - * Example 1: - * - * - * Input: n = 12 - * Output: 3 - * Explanation: 12 = 4 + 4 + 4. - * - * Example 2: - * - * - * Input: n = 13 - * Output: 2 - * Explanation: 13 = 4 + 9. - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/perfect-squares/ -// discuss: https://leetcode.com/problems/perfect-squares/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// BFS -use std::collections::VecDeque; -impl Solution { - pub fn num_squares(n: i32) -> i32 { - if n < 1 { - return 0; - } - let mut deq = VecDeque::new(); - deq.push_back((1, n)); - while let Some((level, num)) = deq.pop_front() { - let mut base = 1; - while base * base <= num { - if base * base == num { - return level; - } - deq.push_back((level + 1, num - base * base)); - base += 1; - } - } - 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_279() { - assert_eq!(Solution::num_squares(13), 2); - assert_eq!(Solution::num_squares(12), 3); - } -} diff --git a/src/solution/s0282_expression_add_operators.rs b/src/solution/s0282_expression_add_operators.rs deleted file mode 100644 index 1c41367b..00000000 --- a/src/solution/s0282_expression_add_operators.rs +++ /dev/null @@ -1,61 +0,0 @@ -/** - * [282] Expression Add Operators - * - * Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value. - * - * Example 1: - * - * - * Input: num = "123", target = 6 - * Output: ["1+2+3", "1*2*3"] - * - * - * Example 2: - * - * - * Input: num = "232", target = 8 - * Output: ["2*3+2", "2+3*2"] - * - * Example 3: - * - * - * Input: num = "105", target = 5 - * Output: ["1*0+5","10-5"] - * - * Example 4: - * - * - * Input: num = "00", target = 0 - * Output: ["0+0", "0-0", "0*0"] - * - * - * Example 5: - * - * - * Input: num = "3456237490", target = 9191 - * Output: [] - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/expression-add-operators/ -// discuss: https://leetcode.com/problems/expression-add-operators/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn add_operators(num: String, target: i32) -> Vec { - vec![] - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_282() {} -} diff --git a/src/solution/s0283_move_zeroes.rs b/src/solution/s0283_move_zeroes.rs deleted file mode 100644 index 6581bcaf..00000000 --- a/src/solution/s0283_move_zeroes.rs +++ /dev/null @@ -1,50 +0,0 @@ -/** - * [283] Move Zeroes - * - * Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. - * - * Example: - * - * - * Input: [0,1,0,3,12] - * Output: [1,3,12,0,0] - * - * Note: - * - *
    - * You must do this in-place without making a copy of the array. - * Minimize the total number of operations. - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/move-zeroes/ -// discuss: https://leetcode.com/problems/move-zeroes/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn move_zeroes(nums: &mut Vec) { - let mut last_none_zero = 0_usize; - for i in 0..nums.len() { - if nums[i] != 0 { - nums.swap(last_none_zero, i); - last_none_zero += 1; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_283() { - let mut vec = vec![0, 1, 0, 3, 12]; - Solution::move_zeroes(&mut vec); - assert_eq!(vec, vec![1, 3, 12, 0, 0]); - } -} diff --git a/src/solution/s0287_find_the_duplicate_number.rs b/src/solution/s0287_find_the_duplicate_number.rs deleted file mode 100644 index cea65001..00000000 --- a/src/solution/s0287_find_the_duplicate_number.rs +++ /dev/null @@ -1,74 +0,0 @@ -/** - * [287] Find the Duplicate Number - * - * Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. - * - * Example 1: - * - * - * Input: [1,3,4,2,2] - * Output: 2 - * - * - * Example 2: - * - * - * Input: [3,1,3,4,2] - * Output: 3 - * - * Note: - * - *
    - * You must not modify the array (assume the array is read only). - * You must use only constant, O(1) extra space. - * Your runtime complexity should be less than O(n^2). - * There is only one duplicate number in the array, but it could be repeated more than once. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-the-duplicate-number/ -// discuss: https://leetcode.com/problems/find-the-duplicate-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// 假如把值看做 next node 的下标, 那么: -// 从 0 出发不会回到 0 -// 一定有环, 因为 1-n 全部落在下标范围 [0, n] 中 -// 从 0 遍历经过的环中, 一定存在重复数字 x, 且 x 就是入环点的下标: -// 1.从 0 走到入环点, 入环点的前驱值为 x; 2.入环点在环上的前驱值也是 x -// 由于我们不可能回到 0, 因此这两个节点下标不同, x 即为要找的重复数字 -impl Solution { - pub fn find_duplicate(nums: Vec) -> i32 { - let mut slow: usize = nums[0] as usize; - let mut fast: usize = nums[nums[0] as usize] as usize; - // util slow meet fast - while slow != fast { - slow = nums[slow] as usize; - fast = nums[nums[fast] as usize] as usize; - } - - fast = 0_usize; - while slow != fast { - fast = nums[fast] as usize; - slow = nums[slow] as usize; - } - slow as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_287() { - assert_eq!(Solution::find_duplicate(vec![1, 3, 4, 2, 2]), 2); - assert_eq!(Solution::find_duplicate(vec![3, 1, 3, 4, 2]), 3); - assert_eq!(Solution::find_duplicate(vec![1, 2, 3, 4, 5, 5]), 5); - assert_eq!(Solution::find_duplicate(vec![5, 1, 2, 3, 4, 5]), 5); - } -} diff --git a/src/solution/s0289_game_of_life.rs b/src/solution/s0289_game_of_life.rs deleted file mode 100644 index 53d34623..00000000 --- a/src/solution/s0289_game_of_life.rs +++ /dev/null @@ -1,118 +0,0 @@ -/** - * [289] Game of Life - * - * According to the Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970." - * - * Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article): - * - *
    - * Any live cell with fewer than two live neighbors dies, as if caused by under-population. - * Any live cell with two or three live neighbors lives on to the next generation. - * Any live cell with more than three live neighbors dies, as if by over-population.. - * Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction. - *
- * - * Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. - * - * Example: - * - * - * Input: - * [ - * [0,1,0], - * [0,0,1], - * [1,1,1], - * [0,0,0] - * ] - * Output: - * [ - * [0,0,0], - * [1,0,1], - * [0,1,1], - * [0,1,0] - * ] - * - * - * Follow up: - * - *
    - * Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells. - * In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems? - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/game-of-life/ -// discuss: https://leetcode.com/problems/game-of-life/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// in-place: 1: live->live, 0: die->die, 2: die->live, 3: live->die -impl Solution { - pub fn game_of_life(board: &mut Vec>) { - let (height, width) = (board.len(), board[0].len()); - let neighbors = vec![ - (-1, -1), - (-1, 0), - (-1, 1), - (0, -1), - (0, 1), - (1, -1), - (1, 0), - (1, 1), - ]; - for i in 0..height { - for j in 0..width { - let mut live = 0; - for offset in neighbors.iter() { - if (offset.0 < 0 && i == 0) - || (offset.0 > 0 && i == height - 1) - || (offset.1 < 0 && j == 0) - || (offset.1 > 0 && j == width - 1) - { - continue; - } - let v = board[(i as i32 + offset.0) as usize][(j as i32 + offset.1) as usize]; - if v == 1 || v == 3 { - live += 1; - } - } - if board[i][j] == 1 && (live < 2 || live > 3) { - // go die - board[i][j] = 3; - } else if board[i][j] == 0 && live == 3 { - // go live - board[i][j] = 2; - } - } - } - - for i in 0..height { - for j in 0..width { - if board[i][j] == 2 { - board[i][j] = 1; - } else if board[i][j] == 3 { - board[i][j] = 0; - } - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_289() { - let mut test = vec![vec![0, 1, 0], vec![0, 0, 1], vec![1, 1, 1], vec![0, 0, 0]]; - Solution::game_of_life(&mut test); - assert_eq!( - test, - vec![vec![0, 0, 0], vec![1, 0, 1], vec![0, 1, 1], vec![0, 1, 0],] - ); - } -} diff --git a/src/solution/s0290_word_pattern.rs b/src/solution/s0290_word_pattern.rs deleted file mode 100644 index c758f352..00000000 --- a/src/solution/s0290_word_pattern.rs +++ /dev/null @@ -1,86 +0,0 @@ -/** - * [290] Word Pattern - * - * Given a pattern and a string str, find if str follows the same pattern. - * - * Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str. - * - * Example 1: - * - * - * Input: pattern = "abba", str = "dog cat cat dog" - * Output: true - * - * Example 2: - * - * - * Input:pattern = "abba", str = "dog cat cat fish" - * Output: false - * - * Example 3: - * - * - * Input: pattern = "aaaa", str = "dog cat cat dog" - * Output: false - * - * Example 4: - * - * - * Input: pattern = "abba", str = "dog dog dog dog" - * Output: false - * - * Notes:
- * You may assume pattern contains only lowercase letters, and str contains lowercase letters that may be separated by a single space. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/word-pattern/ -// discuss: https://leetcode.com/problems/word-pattern/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::HashMap; -impl Solution { - pub fn word_pattern(pattern: String, s: String) -> bool { - let mut p2s = HashMap::new(); - let mut s2p = HashMap::new(); - let mut s_iter = s.split(" "); - let mut p_iter = pattern.chars(); - loop { - match (s_iter.next(), p_iter.next()) { - (Some(sub), Some(ch)) => { - if *p2s.entry(ch).or_insert(sub) != sub || *s2p.entry(sub).or_insert(ch) != ch { - return false; - } - } - (None, None) => break, - _ => return false, - } - } - true - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_290() { - assert_eq!( - Solution::word_pattern("abba".to_owned(), "dog cat cat dog".to_owned()), - true - ); - assert_eq!( - Solution::word_pattern("aaaa".to_owned(), "dog cat cat dog".to_owned()), - false - ); - assert_eq!( - Solution::word_pattern("abba".to_owned(), "dog cat cat fish".to_owned()), - false - ); - } -} diff --git a/src/solution/s0292_nim_game.rs b/src/solution/s0292_nim_game.rs deleted file mode 100644 index 4c46505c..00000000 --- a/src/solution/s0292_nim_game.rs +++ /dev/null @@ -1,40 +0,0 @@ -/** - * [292] Nim Game - * - * You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones. - * - * Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap. - * - * Example: - * - * - * Input: 4 - * Output: false - * Explanation: If there are 4 stones in the heap, then you will never win the game; - * No matter 1, 2, or 3 stones you remove, the last stone will always be - * removed by your friend. - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/nim-game/ -// discuss: https://leetcode.com/problems/nim-game/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn can_win_nim(n: i32) -> bool { - n % 4 != 0 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_292() { - assert_eq!(Solution::can_win_nim(4), false); - } -} diff --git a/src/solution/s0295_find_median_from_data_stream.rs b/src/solution/s0295_find_median_from_data_stream.rs deleted file mode 100644 index 02dc5331..00000000 --- a/src/solution/s0295_find_median_from_data_stream.rs +++ /dev/null @@ -1,136 +0,0 @@ -/** - * [295] Find Median from Data Stream - * - * Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. - * For example, - * - * [2,3,4], the median is 3 - * - * [2,3], the median is (2 + 3) / 2 = 2.5 - * - * Design a data structure that supports the following two operations: - * - * - * void addNum(int num) - Add a integer number from the data stream to the data structure. - * double findMedian() - Return the median of all elements so far. - * - * - * - * - * Example: - * - * - * addNum(1) - * addNum(2) - * findMedian() -> 1.5 - * addNum(3) - * findMedian() -> 2 - * - * - * - * - * Follow up: - * - *
    - * If all integer numbers from the stream are between 0 and 100, how would you optimize it? - * If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it? - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/find-median-from-data-stream/ -// discuss: https://leetcode.com/problems/find-median-from-data-stream/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cmp::Ordering; -use std::collections::BinaryHeap; -#[derive(Eq, PartialEq)] -struct Invert { - value: i32, -} - -impl Ord for Invert { - fn cmp(&self, other: &Invert) -> Ordering { - other.value.cmp(&self.value) - } -} - -impl PartialOrd for Invert { - fn partial_cmp(&self, other: &Invert) -> Option { - Some(self.cmp(other)) - } -} - -struct MedianFinder { - head: BinaryHeap, - tail: BinaryHeap, - count: i32, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl MedianFinder { - /** initialize your data structure here. */ - fn new() -> Self { - MedianFinder { - head: BinaryHeap::new(), - tail: BinaryHeap::new(), - count: 0, - } - } - - fn add_num(&mut self, num: i32) { - self.count += 1; - if self.head.is_empty() || num > self.head.peek().unwrap().value { - self.head.push(Invert { value: num }); - } else { - self.tail.push(num); - } - // balance - if self.head.len() > self.tail.len() + 1 { - self.tail.push(self.head.pop().unwrap().value); - } else if self.head.len() + 1 < self.tail.len() { - self.head.push(Invert { - value: self.tail.pop().unwrap(), - }); - } - } - - fn find_median(&self) -> f64 { - if self.head.len() > self.tail.len() { - self.head.peek().unwrap().value as f64 - } else if self.head.len() < self.tail.len() { - *self.tail.peek().unwrap() as f64 - } else { - (self.head.peek().unwrap().value as f64 + *self.tail.peek().unwrap() as f64) / 2.0 - } - } -} - -/** - * Your MedianFinder object will be instantiated and called as such: - * let obj = MedianFinder::new(); - * obj.add_num(num); - * let ret_2: f64 = obj.find_median(); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_295() { - let mut obj = MedianFinder::new(); - obj.add_num(1); - obj.add_num(2); - assert_eq!(obj.find_median(), 1.5); - obj.add_num(3); - assert_eq!(obj.find_median(), 2_f64); - } -} diff --git a/src/solution/s0299_bulls_and_cows.rs b/src/solution/s0299_bulls_and_cows.rs deleted file mode 100644 index 08525c71..00000000 --- a/src/solution/s0299_bulls_and_cows.rs +++ /dev/null @@ -1,80 +0,0 @@ -/** - * [299] Bulls and Cows - * - * You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number. - * - * Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows. - * - * Please note that both secret number and friend's guess may contain duplicate digits. - * - * Example 1: - * - * - * Input: secret = "1807", guess = "7810" - * - * Output: "1A3B" - * - * Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7. - * - * Example 2: - * - * - * Input: secret = "1123", guess = "0111" - * - * Output: "1A1B" - * - * Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow. - * - * Note: You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal. - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/bulls-and-cows/ -// discuss: https://leetcode.com/problems/bulls-and-cows/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn get_hint(secret: String, guess: String) -> String { - let mut bulls = 0; - let mut cows = 0; - let mut s_iter = secret.chars(); - let mut g_iter = guess.chars(); - let mut bucket = vec![0; 10]; - let mut non_match = vec![]; - while let (Some(s), Some(g)) = (s_iter.next(), g_iter.next()) { - if s == g { - bulls += 1; - } else { - bucket[s.to_digit(10).unwrap() as usize] += 1; - non_match.push(g.to_digit(10).unwrap() as usize); - } - } - for &idx in non_match.iter() { - if bucket[idx] > 0 { - cows += 1; - bucket[idx] -= 1; - } - } - format!("{}A{}B", bulls, cows) - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_299() { - assert_eq!( - Solution::get_hint("1807".to_owned(), "7810".to_owned()), - "1A3B".to_owned() - ); - assert_eq!( - Solution::get_hint("1123".to_owned(), "0111".to_owned()), - "1A1B".to_owned() - ); - } -} diff --git a/src/solution/s0300_longest_increasing_subsequence.rs b/src/solution/s0300_longest_increasing_subsequence.rs deleted file mode 100644 index c0b7eb27..00000000 --- a/src/solution/s0300_longest_increasing_subsequence.rs +++ /dev/null @@ -1,57 +0,0 @@ -/** - * [300] Longest Increasing Subsequence - * - * Given an unsorted array of integers, find the length of longest increasing subsequence. - * - * Example: - * - * - * Input: [10,9,2,5,3,7,101,18] - * Output: 4 - * Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. - * - * Note: - * - * - * There may be more than one LIS combination, it is only necessary for you to return the length. - * Your algorithm should run in O(n^2) complexity. - * - * - * Follow up: Could you improve it to O(n log n) time complexity? - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/longest-increasing-subsequence/ -// discuss: https://leetcode.com/problems/longest-increasing-subsequence/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -// N^2, DP: L[i] = max(1 + L[j]) for j in [0, i) and nums[j] < nums[i] -// N * logN, kick out strategy, maintain an increasing array, new elements kick out a formal one larger than it, if the new element is largest, expand the array -impl Solution { - pub fn length_of_lis(nums: Vec) -> i32 { - let mut incr = Vec::new(); - for &num in nums.iter() { - if let Err(idx) = incr.binary_search(&num) { - if idx >= incr.len() { - incr.push(num); - } else { - incr[idx] = num; - } - } - } - incr.len() as i32 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_300() { - assert_eq!(Solution::length_of_lis(vec![10, 9, 2, 5, 3, 7, 101, 18]), 4); - } -} diff --git a/src/solution/s0301_remove_invalid_parentheses.rs b/src/solution/s0301_remove_invalid_parentheses.rs deleted file mode 100644 index 2cec757c..00000000 --- a/src/solution/s0301_remove_invalid_parentheses.rs +++ /dev/null @@ -1,131 +0,0 @@ -/** - * [301] Remove Invalid Parentheses - * - * Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. - * - * Note: The input string may contain letters other than the parentheses ( and ). - * - * Example 1: - * - * - * Input: "()())()" - * Output: ["()()()", "(())()"] - * - * - * Example 2: - * - * - * Input: "(a)())()" - * Output: ["(a)()()", "(a())()"] - * - * - * Example 3: - * - * - * Input: ")(" - * Output: [""] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/remove-invalid-parentheses/ -// discuss: https://leetcode.com/problems/remove-invalid-parentheses/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// 1. Calculate the number of misplaced left parenthese and right parenthese -// 2. DFS the string to find the all possible removing policy -use std::collections::HashSet; -impl Solution { - pub fn remove_invalid_parentheses(s: String) -> Vec { - let (mut left, mut right) = (0, 0); - let mut chs: Vec = s.chars().collect(); - for &c in chs.iter() { - if c == '(' { - left += 1; - } else if c == ')' { - if left > 0 { - left -= 1; - } else { - right += 1; - } - } - } - - // Now, the number of left and right parentheses are 'left' and 'right' - let mut res: HashSet = HashSet::new(); - let mut seed: Vec = Vec::new(); - Solution::helper(&chs, 0, 0, left, right, &mut seed, &mut res); - res.into_iter().collect() - } - - fn helper( - chs: &Vec, - idx: usize, - left: i32, - l_remain: i32, - r_remain: i32, - exp: &mut Vec, - res: &mut HashSet, - ) { - if idx >= chs.len() { - if left == 0 { - res.insert(exp.iter().collect()); - } - return; - } - if chs[idx] == '(' { - if l_remain > 0 { - Solution::helper( - chs, - idx + 1, - left, - l_remain - 1, - r_remain, - &mut exp.clone(), - res, - ); - } - exp.push('('); - Solution::helper(chs, idx + 1, left + 1, l_remain, r_remain, exp, res); - } else if chs[idx] == ')' { - if r_remain > 0 { - Solution::helper( - chs, - idx + 1, - left, - l_remain, - r_remain - 1, - &mut exp.clone(), - res, - ); - } - if left > 0 { - exp.push(')'); - Solution::helper(chs, idx + 1, left - 1, l_remain, r_remain, exp, res); - } - } else { - exp.push(chs[idx]); - Solution::helper(chs, idx + 1, left, l_remain, r_remain, exp, res); - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_301() { - assert_eq!( - Solution::remove_invalid_parentheses("()())()".to_owned()), - vec_string!["(())()", "()()()"] - ); - assert_eq!( - Solution::remove_invalid_parentheses("(a)())()".to_owned()), - vec_string!["(a)()()", "(a())()"] - ); - } -} diff --git a/src/solution/s0303_range_sum_query_immutable.rs b/src/solution/s0303_range_sum_query_immutable.rs deleted file mode 100644 index c56f8ae9..00000000 --- a/src/solution/s0303_range_sum_query_immutable.rs +++ /dev/null @@ -1,74 +0,0 @@ -/** - * [303] Range Sum Query - Immutable - * - * Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. - * - * Example:
- * - * Given nums = [-2, 0, 3, -5, 2, -1] - * - * sumRange(0, 2) -> 1 - * sumRange(2, 5) -> -1 - * sumRange(0, 5) -> -3 - * - * - * - * Note:
- *
    - * You may assume that the array does not change. - * There are many calls to sumRange function. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/range-sum-query-immutable/ -// discuss: https://leetcode.com/problems/range-sum-query-immutable/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct NumArray { - nums: Vec, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl NumArray { - fn new(nums: Vec) -> Self { - let mut res = 0; - let mut vec = Vec::with_capacity(nums.len()); - for &num in nums.iter() { - res += num; - vec.push(res); - } - NumArray { nums: vec } - } - - fn sum_range(&self, i: i32, j: i32) -> i32 { - let (i, j) = (i as usize, j as usize); - self.nums[j] - if i > 0 { self.nums[i - 1] } else { 0 } - } -} - -/** - * Your NumArray object will be instantiated and called as such: - * let obj = NumArray::new(nums); - * let ret_1: i32 = obj.sum_range(i, j); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_303() { - let mut nums = NumArray::new(vec![-2, 0, 3, -5, 2, -1]); - assert_eq!(nums.sum_range(0, 2), 1); - assert_eq!(nums.sum_range(2, 5), -1); - assert_eq!(nums.sum_range(0, 5), -3); - } -} diff --git a/src/solution/s0304_range_sum_query_2d_immutable.rs b/src/solution/s0304_range_sum_query_2d_immutable.rs deleted file mode 100644 index 42a0f41f..00000000 --- a/src/solution/s0304_range_sum_query_2d_immutable.rs +++ /dev/null @@ -1,124 +0,0 @@ -/** - * [304] Range Sum Query 2D - Immutable - * - * Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). - * - * - * Range Sum Query 2D
- * The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8. - * - * - * Example:
- * - * Given matrix = [ - * [3, 0, 1, 4, 2], - * [5, 6, 3, 2, 1], - * [1, 2, 0, 1, 5], - * [4, 1, 0, 1, 7], - * [1, 0, 3, 0, 5] - * ] - * - * sumRegion(2, 1, 4, 3) -> 8 - * sumRegion(1, 1, 2, 2) -> 11 - * sumRegion(1, 2, 2, 4) -> 12 - * - * - * - * Note:
- *
    - * You may assume that the matrix does not change. - * There are many calls to sumRegion function. - * You may assume that row1 ≤ row2 and col1 ≤ col2. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/range-sum-query-2d-immutable/ -// discuss: https://leetcode.com/problems/range-sum-query-2d-immutable/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct NumMatrix { - cache: Vec>, -} - -/** region[2, 2, 3, 4] = - * x x \ \ \ . square[3,4] - square[1,4] - square[3,1] + square[1,1] - * x x \ \ \ . - * / / o o o . - * / / o o o . - * . . . . . . - * . . . . . . - */ -impl NumMatrix { - fn new(matrix: Vec>) -> Self { - if matrix.is_empty() || matrix[0].is_empty() { - return NumMatrix { - cache: vec![vec![]], - }; - } - let (x_max, y_max) = (matrix.len(), matrix[0].len()); - let mut cache = vec![vec![0; y_max]; x_max]; - for x in 0..x_max { - for y in 0..y_max { - cache[x][y] = matrix[x][y] - + if y > 0 { cache[x][y - 1] } else { 0 } - + if x > 0 { cache[x - 1][y] } else { 0 } - - if x > 0 && y > 0 { - cache[x - 1][y - 1] - } else { - 0 - } - } - } - NumMatrix { cache: cache } - } - - fn sum_region(&self, row1: i32, col1: i32, row2: i32, col2: i32) -> i32 { - let (row1, col1, row2, col2) = (row1 as usize, col1 as usize, row2 as usize, col2 as usize); - self.cache[row2][col2] - - if row1 > 0 { - self.cache[row1 - 1][col2] - } else { - 0 - } - - if col1 > 0 { - self.cache[row2][col1 - 1] - } else { - 0 - } - + if row1 > 0 && col1 > 0 { - self.cache[row1 - 1][col1 - 1] - } else { - 0 - } - } -} - -/** - * Your NumMatrix object will be instantiated and called as such: - * let obj = NumMatrix::new(matrix); - * let ret_1: i32 = obj.sum_region(row1, col1, row2, col2); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_304() { - let matrix = NumMatrix::new(vec![ - vec![3, 0, 1, 4, 2], - vec![5, 6, 3, 2, 1], - vec![1, 2, 0, 1, 5], - vec![4, 1, 0, 1, 7], - vec![1, 0, 3, 0, 5], - ]); - assert_eq!(matrix.sum_region(1, 1, 2, 2), 11); - assert_eq!(matrix.sum_region(2, 1, 4, 3), 8); - assert_eq!(matrix.sum_region(1, 2, 2, 4), 12); - } -} diff --git a/src/solution/s0306_additive_number.rs b/src/solution/s0306_additive_number.rs deleted file mode 100644 index 38f6c0b2..00000000 --- a/src/solution/s0306_additive_number.rs +++ /dev/null @@ -1,113 +0,0 @@ -/** - * [306] Additive Number - * - * Additive number is a string whose digits can form additive sequence. - * - * A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two. - * - * Given a string containing only digits '0'-'9', write a function to determine if it's an additive number. - * - * Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid. - * - * Example 1: - * - * - * Input: "112358" - * Output: true - * Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. - * 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8 - * - * - * Example 2: - * - * - * Input: "199100199" - * Output: true - * Explanation: The additive sequence is: 1, 99, 100, 199. - * 1 + 99 = 100, 99 + 100 = 199 - * - * Follow up:
- * How would you handle overflow for very large input integers? - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/additive-number/ -// discuss: https://leetcode.com/problems/additive-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -// first_cut second_cut third_cut -// V V V -// 1 99 100 199 -impl Solution { - pub fn is_additive_number(num: String) -> bool { - let mut chs: Vec = num.chars().map(|c| c.to_digit(10).unwrap()).collect(); - let mut num1 = 0; - let len = chs.len(); - // first cut - for i in 0..(len / 2 + 1) { - num1 = num1 * 10 + chs[i]; - if Solution::second_cut(i + 1, len, num1, &chs) { - return true; - } - if num1 == 0 { - break; - } - } - false - } - - fn second_cut(from: usize, len: usize, num1: u32, chs: &Vec) -> bool { - let mut num2 = 0; - for i in from..len { - num2 = num2 * 10 + chs[i]; - if Solution::third_cut(i + 1, len, num1, num2, chs, false) { - return true; - } - if num2 == 0 { - break; - } - } - false - } - - fn third_cut( - from: usize, - len: usize, - num1: u32, - num2: u32, - chs: &Vec, - found: bool, - ) -> bool { - if found && from >= len { - return true; - } - let mut num3 = 0; - for i in from..len { - num3 = num3 * 10 + chs[i]; - if num3 == num2 + num1 { - if Solution::third_cut(i + 1, len, num2, num3, chs, true) { - return true; - } - } else if num3 == 0 || num3 > num1 + num2 { - break; - } - } - false - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_306() { - assert_eq!(Solution::is_additive_number("112358".to_owned()), true); - assert_eq!(Solution::is_additive_number("199100199".to_owned()), true); - assert_eq!(Solution::is_additive_number("1991001990".to_owned()), false); - assert_eq!(Solution::is_additive_number("1023".to_owned()), false); - } -} diff --git a/src/solution/s0307_range_sum_query_mutable.rs b/src/solution/s0307_range_sum_query_mutable.rs deleted file mode 100644 index bbaac119..00000000 --- a/src/solution/s0307_range_sum_query_mutable.rs +++ /dev/null @@ -1,151 +0,0 @@ -/** - * [307] Range Sum Query - Mutable - * - * Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. - * - * The update(i, val) function modifies nums by updating the element at index i to val. - * - * Example: - * - * - * Given nums = [1, 3, 5] - * - * sumRange(0, 2) -> 9 - * update(1, 2) - * sumRange(0, 2) -> 8 - * - * - * Note: - * - *
    - * The array is only modifiable by the update function. - * You may assume the number of calls to update and sumRange function is distributed evenly. - *
- * - */ -pub struct Solution {} - -// Segement Tree -// -// N[0:6] -// / \ -// / \ -// N[0:3] N[4:6] -// / \ / \ -// N[0:1] N[2:3] N[4:5] N[6] -// / \ / \ / \ -// N[0] N[1] N[2] N[3] N[4] N[5] -// -// problem: https://leetcode.com/problems/range-sum-query-mutable/ -// discuss: https://leetcode.com/problems/range-sum-query-mutable/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -struct NumArray { - tree: Vec, - n: usize, -} - -/** - * `&self` means the method takes an immutable reference. - * If you need a mutable reference, change it to `&mut self` instead. - */ -impl NumArray { - fn new(nums: Vec) -> Self { - let n = nums.len(); - let mut tree = vec![0; 4 * n]; - if n > 0 { - NumArray::build(1, 0, n - 1, &mut tree, &nums); - } - NumArray { tree: tree, n: n } - } - - fn update(&mut self, i: i32, val: i32) { - NumArray::update_internal(i as usize, val, 1, 0, self.n - 1, &mut self.tree); - } - - fn sum_range(&self, i: i32, j: i32) -> i32 { - NumArray::sum(1, 0, self.n - 1, i as usize, j as usize, &self.tree) - } - - fn build(node: usize, start: usize, end: usize, tree: &mut Vec, nums: &Vec) { - if start == end { - tree[node] = nums[start]; - } else { - let mid = (start + end) / 2; - NumArray::build(2 * node, start, mid, tree, nums); - NumArray::build(2 * node + 1, mid + 1, end, tree, nums); - tree[node] = tree[2 * node] + tree[2 * node + 1]; - } - } - - fn update_internal( - i: usize, - val: i32, - node: usize, - start: usize, - end: usize, - tree: &mut Vec, - ) { - if start == end { - tree[node] = val; - } else { - let mid = (start + end) / 2; - if i <= mid { - NumArray::update_internal(i, val, 2 * node, start, mid, tree); - } else { - NumArray::update_internal(i, val, 2 * node + 1, mid + 1, end, tree); - } - tree[node] = tree[2 * node] + tree[2 * node + 1]; - } - } - - fn sum( - node: usize, - start: usize, - end: usize, - left: usize, - right: usize, - tree: &Vec, - ) -> i32 { - if right < start || left > end { - // not in range - 0 - } else if left <= start && end <= right { - // completely in range - tree[node] - } else { - // partially in range - let mid = (start + end) / 2; - NumArray::sum(2 * node, start, mid, left, right, tree) - + NumArray::sum(2 * node + 1, mid + 1, end, left, right, tree) - } - } -} - -/** - * Your NumArray object will be instantiated and called as such: - * let obj = NumArray::new(nums); - * obj.update(i, val); - * let ret_2: i32 = obj.sum_range(i, j); - */ - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_307() { - let _empty = NumArray::new(vec![]); - let mut tree = NumArray::new(vec![1, 1, 1, 1, 1, 1, 1, 1, 1, 1]); - assert_eq!(tree.sum_range(0, 6), 7); - tree.update(0, 2); - assert_eq!(tree.sum_range(0, 6), 8); - tree.update(1, 2); - assert_eq!(tree.sum_range(0, 2), 5); - tree.update(6, 10); - assert_eq!(tree.sum_range(6, 6), 10); - } -} diff --git a/src/solution/s0309_best_time_to_buy_and_sell_stock_with_cooldown.rs b/src/solution/s0309_best_time_to_buy_and_sell_stock_with_cooldown.rs deleted file mode 100644 index 6f3e911d..00000000 --- a/src/solution/s0309_best_time_to_buy_and_sell_stock_with_cooldown.rs +++ /dev/null @@ -1,79 +0,0 @@ -/** - * [309] Best Time to Buy and Sell Stock with Cooldown - * - * Say you have an array for which the i^th element is the price of a given stock on day i. - * - * Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions: - * - * - * You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). - * After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day) - * - * - * Example: - * - * - * Input: [1,2,3,0,2] - * Output: 3 - * Explanation: transactions = [buy, sell, cooldown, buy, sell] - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/ -// discuss: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -dp[i]: max profit with selling at day i -dp2[i]: max profit till day i - -dp[i] = max(dp[i-1] + p[i] - p[i-1], dp2[i-2], dp2[i-3] + p[i] - p[i-1]) -*/ -impl Solution { - pub fn max_profit(prices: Vec) -> i32 { - if prices.len() < 2 { - return 0; - } - if prices.len() == 2 { - return i32::max(0, prices[1] - prices[0]); - } - let mut dp = vec![0; prices.len()]; - let mut dp2 = vec![0; prices.len()]; - let mut max = 0; - dp[0] = 0; - dp2[0] = 0; - dp[1] = prices[1] - prices[0]; - dp2[1] = i32::max(dp2[0], dp[1]); - dp[2] = i32::max(prices[2] - prices[1], prices[2] - prices[0]); - dp2[2] = i32::max(dp2[1], dp[2]); - for i in 3..prices.len() { - dp[i] = i32::max( - dp[i - 1] + prices[i] - prices[i - 1], - i32::max(dp2[i - 2], dp2[i - 3] + prices[i] - prices[i - 1]), - ); - dp2[i] = i32::max(dp2[i - 1], dp[i]); - } - let mut temp = 0; - for &m in dp.iter() { - if m > temp { - temp = m; - } - } - temp - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_309() { - assert_eq!(Solution::max_profit(vec![1, 2, 3, 0, 2]), 3); - assert_eq!(Solution::max_profit(vec![4, 2, 7, 1, 11]), 10); - } -} diff --git a/src/solution/s0310_minimum_height_trees.rs b/src/solution/s0310_minimum_height_trees.rs deleted file mode 100644 index 4141273c..00000000 --- a/src/solution/s0310_minimum_height_trees.rs +++ /dev/null @@ -1,110 +0,0 @@ -/** - * [310] Minimum Height Trees - * - * For an undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels. - * - * Format
- * The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels). - * - * You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges. - * - * Example 1 : - * - * - * Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]] - * - * 0 - * | - * 1 - * / \ - * 2 3 - * - * Output: [1] - * - * - * Example 2 : - * - * - * Input: n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]] - * - * 0 1 2 - * \ | / - * 3 - * | - * 4 - * | - * 5 - * - * Output: [3, 4] - * - * Note: - * - * - * According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.” - * The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/minimum-height-trees/ -// discuss: https://leetcode.com/problems/minimum-height-trees/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::mem; -impl Solution { - pub fn find_min_height_trees(n: i32, edges: Vec>) -> Vec { - let n = n as usize; - let mut matrix: Vec> = vec![vec![]; n]; - for edge in edges.iter() { - matrix[edge[0] as usize].push(edge[1] as usize); - matrix[edge[1] as usize].push(edge[0] as usize); - } - let mut count = n; - let mut la: Vec = vec![]; - let mut lb: Vec = vec![]; - for i in 0..n { - if matrix[i].len() <= 1 { - la.push(i); - } - } - while count > 2 { - count -= la.len(); - for &i in la.iter() { - let j = matrix[i][0]; - let idx = matrix[j].iter().position(|&r| r == i).unwrap(); - matrix[j].remove(idx); - if matrix[j].len() == 1 { - lb.push(j); - } - } - la.clear(); - mem::swap(&mut la, &mut lb); - } - la.into_iter().map(|i| i as i32).collect() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_310() { - assert_eq!( - Solution::find_min_height_trees(4, vec![vec![1, 0], vec![1, 2], vec![1, 3]]), - vec![1] - ); - assert_eq!( - Solution::find_min_height_trees( - 6, - vec![vec![0, 3], vec![1, 3], vec![2, 3], vec![4, 3], vec![5, 4]] - ), - vec![3, 4] - ); - assert_eq!(Solution::find_min_height_trees(1, vec![]), vec![0]); - } -} diff --git a/src/solution/s0312_burst_balloons.rs b/src/solution/s0312_burst_balloons.rs deleted file mode 100644 index f8833b5e..00000000 --- a/src/solution/s0312_burst_balloons.rs +++ /dev/null @@ -1,108 +0,0 @@ -/** - * [312] Burst Balloons - * - * Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent. - * - * Find the maximum coins you can collect by bursting the balloons wisely. - * - * Note: - * - * - * You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them. - * 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100 - * - * - * Example: - * - * - * Input: [3,1,5,8] - * Output: 167 - * Explanation: nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> [] - * coins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167 - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/burst-balloons/ -// discuss: https://leetcode.com/problems/burst-balloons/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -/* -The key idea is, for a sequence of balloon, select a balloon to be the last one to be bursted: - -max of [1 . a b c d e f . 1] - - ^ say we select 'c' as the last balloon to burst, then: - -= - max of [1 . a b . c] + - - max of [c . d e f . 1] + - - 1 * c * 1 - -Then we can use memorize to record the max of every sub sequence -*/ -impl Solution { - pub fn max_coins(nums: Vec) -> i32 { - if nums.is_empty() { - return 0; - } - let mut coins = vec![0; nums.len() + 2]; - let mut len = 0_usize; - // filter out zeros - for &num in nums.iter() { - if num != 0 { - len += 1; - coins[len] = num; - } - } - coins[0] = 1; - coins[len + 1] = 1; - - let mut memo = vec![vec![0; len + 1]; len + 1]; - Solution::max_subrange(&coins, 1, len, &mut memo) - } - - fn max_subrange(coins: &Vec, start: usize, end: usize, memo: &mut Vec>) -> i32 { - if memo[start][end] != 0 { - return memo[start][end]; - } - if start == end { - memo[start][end] = coins[start - 1] * coins[start] * coins[start + 1]; - return memo[start][end]; - } - let mut max = 0; - for i in start..end + 1 { - let left_max = if i > start { - Solution::max_subrange(coins, start, i - 1, memo) - } else { - 0 - }; - let right_max = if i < end { - Solution::max_subrange(coins, i + 1, end, memo) - } else { - 0 - }; - max = i32::max( - max, - left_max + right_max + coins[i] * coins[start - 1] * coins[end + 1], - ); - } - memo[start][end] = max; - return memo[start][end]; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_312() { - assert_eq!(Solution::max_coins(vec![3, 1, 5, 8]), 167); - } -} diff --git a/src/solution/s0313_super_ugly_number.rs b/src/solution/s0313_super_ugly_number.rs deleted file mode 100644 index bf99bcc8..00000000 --- a/src/solution/s0313_super_ugly_number.rs +++ /dev/null @@ -1,94 +0,0 @@ -/** - * [313] Super Ugly Number - * - * Write a program to find the n^th super ugly number. - * - * Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. - * - * Example: - * - * - * Input: n = 12, primes = [2,7,13,19] - * Output: 32 - * Explanation: [1,2,4,7,8,13,14,16,19,26,28,32] is the sequence of the first 12 - * super ugly numbers given primes = [2,7,13,19] of size 4. - * - * Note: - * - * - * 1 is a super ugly number for any given primes. - * The given numbers in primes are in ascending order. - * 0 < k ≤ 100, 0 < n ≤ 10^6, 0 < primes[i] < 1000. - * The n^th super ugly number is guaranteed to fit in a 32-bit signed integer. - * - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/super-ugly-number/ -// discuss: https://leetcode.com/problems/super-ugly-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::cmp::Ordering; -use std::collections::BinaryHeap; -#[derive(Eq, PartialEq)] -struct Invert { - base: i32, - idx: usize, - value: i32, -} - -impl Ord for Invert { - fn cmp(&self, other: &Invert) -> Ordering { - other.value.cmp(&self.value) - } -} - -impl PartialOrd for Invert { - fn partial_cmp(&self, other: &Invert) -> Option { - Some(self.cmp(other)) - } -} - -impl Solution { - pub fn nth_super_ugly_number(n: i32, primes: Vec) -> i32 { - let mut vec = vec![1; 1]; - let mut heap: BinaryHeap = BinaryHeap::new(); - for &prime in primes.iter() { - heap.push(Invert { - base: prime, - idx: 0, - value: prime, - }); - } - for _ in 0..n - 1 { - let mut min = 0; - if let Some(num) = heap.peek() { - min = num.value; - } - vec.push(min); - while heap.peek().unwrap().value == min { - let p = heap.pop().unwrap(); - heap.push(Invert { - base: p.base, - idx: p.idx + 1, - value: p.base * vec[p.idx + 1], - }); - } - } - *vec.last().unwrap() - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_313() { - assert_eq!(Solution::nth_super_ugly_number(12, vec![2, 7, 13, 19]), 32); - } -} diff --git a/src/solution/s0509_fibonacci_number.rs b/src/solution/s0509_fibonacci_number.rs deleted file mode 100644 index 47a9b420..00000000 --- a/src/solution/s0509_fibonacci_number.rs +++ /dev/null @@ -1,78 +0,0 @@ -/** - * [1013] Fibonacci Number - * - * The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is, - * - * - * F(0) = 0, F(1) = 1 - * F(N) = F(N - 1) + F(N - 2), for N > 1. - * - * - * Given N, calculate F(N). - * - * - * - * Example 1: - * - * - * Input: 2 - * Output: 1 - * Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1. - * - * - * Example 2: - * - * - * Input: 3 - * Output: 2 - * Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2. - * - * - * Example 3: - * - * - * Input: 4 - * Output: 3 - * Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3. - * - * - * - * - * Note: - * - * 0 ≤ N ≤ 30. - * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/fibonacci-number/ -// discuss: https://leetcode.com/problems/fibonacci-number/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn fib(n: i32) -> i32 { - if n == 0 { - return 0; - } - let mut f = (0, 1); - for _ in 1..n { - f = (f.1, f.0 + f.1); - } - return f.1; - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_1013() { - assert_eq!(Solution::fib(2), 1); - assert_eq!(Solution::fib(3), 2); - assert_eq!(Solution::fib(4), 3); - } -} diff --git a/src/solution/s0704_binary_search.rs b/src/solution/s0704_binary_search.rs deleted file mode 100644 index 26ed0c8d..00000000 --- a/src/solution/s0704_binary_search.rs +++ /dev/null @@ -1,83 +0,0 @@ -/** - * [792] Binary Search - * - * Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1. - * - *
- * Example 1: - * - * - * Input: nums = [-1,0,3,5,9,12], target = 9 - * Output: 4 - * Explanation: 9 exists in nums and its index is 4 - * - * - * - * Example 2: - * - * - * Input: nums = [-1,0,3,5,9,12], target = 2 - * Output: -1 - * Explanation: 2 does not exist in nums so return -1 - * - * - * - * - * Note: - * - *
    - * You may assume that all elements in nums are unique. - * n will be in the range [1, 10000]. - * The value of each element in nums will be in the range [-9999, 9999]. - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/binary-search/ -// discuss: https://leetcode.com/problems/binary-search/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here -use std::cmp::Ordering; - -impl Solution { - pub fn search(nums: Vec, target: i32) -> i32 { - let mut lo = 0i32; - let mut hi = (nums.len() as i32) - 1; - while lo <= hi { - let mid = lo + (hi - lo) / 2; - match nums[mid as usize].cmp(&target) { - Ordering::Less => { - lo = mid + 1; - } - Ordering::Greater => { - hi = mid - 1; - } - Ordering::Equal => { - return mid; - } - } - } - -1 - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_792() { - assert_eq!(Solution::search(vec![-1, 0, 3, 5, 9, 12], 9), 4); - assert_eq!(Solution::search(vec![-1, 0, 3, 5, 9, 12], 2), -1); - assert_eq!(Solution::search(vec![1], 1), 0); - assert_eq!(Solution::search(vec![5], -5), -1); - assert_eq!(Solution::search(vec![5], 6), -1); - assert_eq!(Solution::search(vec![1, 2], 0), -1); - assert_eq!(Solution::search(vec![1, 2], 1), 0); - assert_eq!(Solution::search(vec![1, 2], 2), 1); - assert_eq!(Solution::search(vec![1, 2], 3), -1); - } -} diff --git a/src/solution/s0969_pancake_sorting.rs b/src/solution/s0969_pancake_sorting.rs deleted file mode 100644 index 541ae423..00000000 --- a/src/solution/s0969_pancake_sorting.rs +++ /dev/null @@ -1,148 +0,0 @@ -/** - * [1009] Pancake Sorting - * - * Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A. We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A. - * - * Return the k-values corresponding to a sequence of pancake flips that sort A. Any valid answer that sorts the array within 10 * A.length flips will be judged as correct. - * - * - * - * Example 1: - * - * - * Input: [3,2,4,1] - * Output: [4,2,4,3] - * Explanation: - * We perform 4 pancake flips, with k values 4, 2, 4, and 3. - * Starting state: A = [3, 2, 4, 1] - * After 1st flip (k=4): A = [1, 4, 2, 3] - * After 2nd flip (k=2): A = [4, 1, 2, 3] - * After 3rd flip (k=4): A = [3, 2, 1, 4] - * After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted. - * - * - *
- * Example 2: - * - * - * Input: [1,2,3] - * Output: [] - * Explanation: The input is already sorted, so there is no need to flip anything. - * Note that other answers, such as [3, 3], would also be accepted. - * - * - * - *
- * - * Note: - * - *
    - * 1 <= A.length <= 100 - * A[i] is a permutation of [1, 2, ..., A.length] - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/pancake-sorting/ -// discuss: https://leetcode.com/problems/pancake-sorting/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn pancake_sort(a: Vec) -> Vec { - let len = a.len(); - if len <= 1 { - return Vec::new(); - } - let mut b = a.clone(); - let mut res: Vec = Vec::new(); - for i in 0..len { - if i == (len - 1) { - break; - } - let k = (len - i) as i32; - let index = Solution::find_k(&b, k); - if index == (k - 1) as usize { - continue; - } - if index != 0usize { - Solution::pancake_oper(&mut b, index, &mut res); - } - Solution::pancake_oper(&mut b, (k - 1) as usize, &mut res); - } - // println!("{:?}", b); - res - } - - fn find_k(a: &Vec, k: i32) -> usize { - for i in 0..(k - 1) { - if a[i as usize] == k { - return i as usize; - } - } - (k - 1) as usize - } - - pub fn pancake_oper(a: &mut Vec, index: usize, res: &mut Vec) { - let mut helper = Vec::new(); - for i in 0..(index + 1) { - helper.push(a[index - i]); - } - for i in 0..(index + 1) { - a[i] = helper[i]; - } - res.push((index + 1) as i32); - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - use rand::seq::SliceRandom; - use rand::{thread_rng, Rng}; - - #[test] - fn test_1009() { - for _i in 0..20 { - let mut rng = rand::thread_rng(); - let size = rng.gen_range(0, 1000); - let sorted_vector = make_sorted_vector(size); - let mut shuffled_vector = make_shuffled_vector(&sorted_vector); - let res = Solution::pancake_sort(shuffled_vector.clone()); - let oper_num = res.len(); - apply_pancake_sort_res(&mut shuffled_vector, res); - assert_eq!(shuffled_vector, sorted_vector); - assert!(oper_num < (size * 10) as usize); - } - } - - fn make_sorted_vector(i: i32) -> Vec { - (1..i + 1).collect() - } - - fn make_shuffled_vector(a: &Vec) -> Vec { - let mut rng = thread_rng(); - let mut res = a.clone(); - res.shuffle(&mut rng); - res - } - - fn apply_pancake_sort_res(shuffled_vecter: &mut Vec, oper: Vec) { - for i in oper { - pancake_oper(shuffled_vecter, (i - 1) as usize); - } - } - - pub fn pancake_oper(a: &mut Vec, index: usize) { - let mut helper = Vec::new(); - for i in 0..(index + 1) { - helper.push(a[index - i]); - } - for i in 0..(index + 1) { - a[i] = helper[i]; - } - } -} diff --git a/src/solution/s1018_binary_prefix_divisible_by_5.rs b/src/solution/s1018_binary_prefix_divisible_by_5.rs deleted file mode 100644 index b2bb1839..00000000 --- a/src/solution/s1018_binary_prefix_divisible_by_5.rs +++ /dev/null @@ -1,106 +0,0 @@ -/** - * [1071] Binary Prefix Divisible By 5 - * - * Given an array A of 0s and 1s, consider N_i: the i-th subarray from A[0] to A[i] interpreted as a binary number (from most-significant-bit to least-significant-bit.) - * - * Return a list of booleans answer, where answer[i] is true if and only if N_i is divisible by 5. - * - * Example 1: - * - * - * Input: [0,1,1] - * Output: [true,false,false] - * Explanation: - * The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10. Only the first number is divisible by 5, so answer[0] is true. - * - * - * Example 2: - * - * - * Input: [1,1,1] - * Output: [false,false,false] - * - * - * Example 3: - * - * - * Input: [0,1,1,1,1,1] - * Output: [true,false,false,false,true,false] - * - * - * Example 4: - * - * - * Input: [1,1,1,0,1] - * Output: [false,false,false,false,false] - * - * - * - * - * Note: - * - *
    - * 1 <= A.length <= 30000 - * A[i] is 0 or 1 - *
- * - */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/binary-prefix-divisible-by-5/ -// discuss: https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -impl Solution { - pub fn prefixes_div_by5(a: Vec) -> Vec { - let mut ret = vec![]; - let mut n = 0; - for i in a { - let remain = (n * 2 + i) % 5; - ret.push(remain == 0); - n = remain; - } - - ret - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_1071() { - assert_eq!( - Solution::prefixes_div_by5(vec![0, 1, 1]), - vec![true, false, false] - ); - assert_eq!( - Solution::prefixes_div_by5(vec![1, 1, 1]), - vec![false, false, false] - ); - assert_eq!( - Solution::prefixes_div_by5(vec![0, 1, 1, 1, 1, 1]), - vec![true, false, false, false, true, false] - ); - assert_eq!( - Solution::prefixes_div_by5(vec![1, 1, 1, 0, 1]), - vec![false, false, false, false, false] - ); - assert_eq!( - Solution::prefixes_div_by5(vec![ - 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, - 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1 - ]), - vec![ - false, false, false, false, false, false, false, false, false, false, false, false, - false, false, false, false, false, false, false, false, false, false, false, false, - false, false, false, false, false, false, false, true, false, false, true, true, - true, true, false - ] - ); - } -} diff --git a/src/solution/s1046_last_stone_weight.rs b/src/solution/s1046_last_stone_weight.rs deleted file mode 100644 index 871b4eb4..00000000 --- a/src/solution/s1046_last_stone_weight.rs +++ /dev/null @@ -1,78 +0,0 @@ -/** - * [1127] Last Stone Weight - * - * We have a collection of rocks, each rock has a positive integer weight. - * - * Each turn, we choose the two heaviest rocks and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is: - * - * - * If x == y, both stones are totally destroyed; - * If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x. - * - * - * At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.) - * - * - * - * Example 1: - * - * - * Input: [2,7,4,1,8,1] - * Output: 1 - * Explanation: - * We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then, - * we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then, - * we combine 2 and 1 to get 1 so the array converts to [1,1,1] then, - * we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone. - * - * - * - * Note: - * - *
    - * 1 <= stones.length <= 30 - * 1 <= stones[i] <= 1000 - *
- */ -pub struct Solution {} - -// problem: https://leetcode.com/problems/last-stone-weight/ -// discuss: https://leetcode.com/problems/last-stone-weight/discuss/?currentPage=1&orderBy=most_votes&query= - -// submission codes start here - -use std::collections::BinaryHeap; -impl Solution { - pub fn last_stone_weight(stones: Vec) -> i32 { - let mut heap = BinaryHeap::new(); - heap.extend(stones); - loop { - if let Some(rock1) = heap.pop() { - if let Some(rock2) = heap.pop() { - if rock1 > rock2 { - heap.push(rock1 - rock2); - } - } else { - return rock1; - } - } else { - return 0; - } - } - } -} - -// submission codes end - -#[cfg(test)] -mod tests { - use super::*; - - #[test] - fn test_1127() { - assert_eq!(Solution::last_stone_weight(vec![2, 7, 4, 1, 8, 1]), 1); - assert_eq!(Solution::last_stone_weight(vec![2]), 2); - assert_eq!(Solution::last_stone_weight(vec![2, 2]), 0); - assert_eq!(Solution::last_stone_weight(vec![1, 2, 2]), 1); - } -} From 7431924d2e8dd225912d92bcc197039ac6c11adc Mon Sep 17 00:00:00 2001 From: pluveto Date: Sun, 15 Jan 2023 02:25:03 +0000 Subject: [PATCH 2/2] Get hot problems --- FetchList.js | 86 +++++ HOT.md | 995 +++++++++++++++++++++++++++++++++++++++++++++++++++ README.md | 2 + 3 files changed, 1083 insertions(+) create mode 100644 FetchList.js create mode 100644 HOT.md diff --git a/FetchList.js b/FetchList.js new file mode 100644 index 00000000..cfd1bc42 --- /dev/null +++ b/FetchList.js @@ -0,0 +1,86 @@ +// https://codetop.cc/api/questions/?page=${N}&search=&ordering=-frequency + +let genUrl = (N) => { + return `https://codetop.cc/api/questions/?page=${N}&search=&ordering=-frequency` +} + +let pageRange = {min: 1, max: 50} + +let fetchPage = (N) => { + let url = genUrl(N) + console.log(`Fetching ${url}...`); + return fetch(url).then(r => r.json()) +} + +let fetchAllPages = (range) => { + let queryPeriod = 1000 + let promises = [] + for (let i = range.min; i <= range.max; i++) { + promises.push(fetchPage(i)) + } + return Promise.all(promises) +} + +let fetchAllQuestions = () => { + return fetchAllPages(pageRange).then(pages => { + let questions = [] + for (let page of pages) { + console.log("Page: ", page); + questions = questions.concat(page.list) + } + return questions + }) +} + +/** + * Response format: + * { + * count, + * list { + * id, + * value, + * leetcode? { + * id, + * title + * } + * } + * } + */ + +// Generate a list of {leet_id, value, title} +let genList = (questions) => { + let list = [] + for (let q of questions) { + let leetcode = q.leetcode + if (leetcode) { + let id = leetcode.id + let title = leetcode.title + let value = q.value + list.push({id, value, title}) + } + } + return list +} + +// Sort the list by value +let sortList = (list) => { + return list.sort((a, b) => { + return b.value - a.value + }) +} + +// Print to console +let printList = (list) => { + let out = "" + for (let {id, value, title} of list) { + out += `${id},${value},${title}\n` + } + console.log(out) +} + +// Main +fetchAllQuestions().then(questions => { + let list = genList(questions) + let sortedList = sortList(list) + printList(sortedList) +}) \ No newline at end of file diff --git a/HOT.md b/HOT.md new file mode 100644 index 00000000..e7b518bd --- /dev/null +++ b/HOT.md @@ -0,0 +1,995 @@ +- [ ] LC 1681 F545 无重复字符的最长子串 +- [ ] LC 1478 F532 反转链表 +- [ ] LC 1538 F432 LRU缓存机制 +- [ ] LC 1469 F380 数组中的第K个最大元素 +- [ ] LC 1659 F287 K 个一组翻转链表 +- [ ] LC 1669 F267 三数之和 +- [ ] LC 1631 F237 最大子序和 +- [ ] LC 1906 F234 手撕快速排序 +- [ ] LC 1663 F221 合并两个有序链表 +- [ ] LC 1683 F212 两数之和 +- [ ] LC 1582 F202 二叉树的层序遍历 +- [ ] LC 1664 F194 有效的括号 +- [ ] LC 1651 F193 搜索旋转排序数组 +- [ ] LC 1563 F191 买卖股票的最佳时机 +- [ ] LC 1484 F189 岛屿数量 +- [ ] LC 1679 F188 最长回文子串 +- [ ] LC 1543 F188 环形链表 +- [ ] LC 1596 F185 合并两个有序数组 +- [ ] LC 1581 F183 二叉树的锯齿形层次遍历 +- [ ] LC 1448 F183 二叉树的最近公共祖先 +- [ ] LC 1638 F181 全排列 +- [ ] LC 1524 F170 相交链表 +- [ ] LC 1630 F161 螺旋矩阵 +- [ ] LC 1661 F158 合并K个排序链表 +- [ ] LC 1592 F153 反转链表 II +- [ ] LC 1269 F148 字符串相加 +- [ ] LC 1542 F144 环形链表 II +- [ ] LC 1384 F138 最长上升子序列 +- [ ] LC 1642 F137 接雨水 +- [ ] LC 1541 F122 重排链表 +- [ ] LC 1560 F122 二叉树中的最大路径和 +- [ ] LC 1665 F115 删除链表的倒数第N个节点 +- [ ] LC 1590 F115 二叉树的中序遍历 +- [ ] LC 1612 F114 编辑距离 +- [ ] LC 925 F114 二分查找 +- [ ] LC 1680 F113 寻找两个正序数组的中位数 +- [ ] LC 1452 F112 用栈实现队列 +- [ ] LC 1485 F111 二叉树的右视图 +- [ ] LC 1614 F106 爬楼梯 +- [ ] LC 1628 F106 合并区间 +- [ ] LC 1536 F105 排序链表 +- [ ] LC 1653 F104 下一个排列 +- [ ] LC 1602 F104 删除排序链表中的重复元素 II +- [ ] LC 1615 F101 x 的平方根 +- [ ] LC 1682 F100 两数相加 +- [ ] LC 485 F98 最长公共子序列 +- [ ] LC 1676 F98 字符串转换整数 (atoi) +- [ ] LC 1591 F95 复原IP地址 +- [ ] LC 1662 F94 括号生成 +- [ ] LC 1445 F93 滑动窗口最大值 +- [ ] LC 1643 F91 缺失的第一个正数 +- [ ] LC 122 F91 链表中倒数第k个节点 +- [ ] LC 1608 F80 最小覆盖子串 +- [ ] LC 1579 F80 从前序与中序遍历序列构造二叉树 +- [ ] LC 1362 F79 零钱兑换 +- [ ] LC 1519 F79 比较版本号 +- [ ] LC 1533 F78 翻转字符串里的单词 +- [ ] LC 1540 F77 二叉树的前序遍历 +- [ ] LC 1641 F76 字符串相乘 +- [ ] LC 1606 F75 子集 +- [ ] LC 1574 F75 平衡二叉树 +- [ ] LC 1652 F73 最长有效括号 +- [ ] LC 1580 F73 二叉树的最大深度 +- [ ] LC 1529 F73 最小栈 +- [ ] LC 1555 F72 求根到叶子节点数字之和 +- [ ] LC 1583 F70 对称二叉树 +- [ ] LC 1156 F69 二叉树的直径 +- [ ] LC 1586 F68 验证二叉搜索树 +- [ ] LC 1620 F68 最小路径和 +- [ ] LC 821 F67 用 Rand7() 实现 Rand10() +- [ ] LC 1571 F67 路径总和 II +- [ ] LC 1572 F63 路径总和 +- [ ] LC 1636 F63 旋转图像 +- [ ] LC 1645 F62 组合总和 +- [ ] LC 1450 F61 回文链表 +- [ ] LC 1515 F60 多数元素 +- [ ] LC 999 F58 最长重复子数组 +- [ ] LC 1463 F57 最大正方形 +- [ ] LC 1670 F56 最长公共前缀 +- [ ] LC 1650 F56 在排序数组中查找元素的第一个和最后一个位置 +- [ ] LC 1290 F56 字符串解码 +- [ ] LC 1444 F56 搜索二维矩阵 II +- [ ] LC 1458 F56 翻转二叉树 +- [ ] LC 1556 F53 最长连续序列 +- [ ] LC 1522 F53 寻找峰值 +- [ ] LC 1011 F52 岛屿的最大面积 +- [ ] LC 1622 F51 不同路径 +- [ ] LC 1457 F51 基本计算器 II +- [ ] LC 1601 F51 删除排序链表中的重复元素 +- [ ] LC 1562 F49 买卖股票的最佳时机 II +- [ ] LC 1486 F49 打家劫舍 +- [ ] LC 1505 F48 最大数 +- [ ] LC 1908 F48 手撕堆排序 +- [ ] LC 1660 F48 两两交换链表中的节点 +- [ ] LC 1532 F47 乘积最大子数组 +- [ ] LC 1401 F47 移动零 +- [ ] LC 1044 F47 二叉树最大宽度 +- [ ] LC 1546 F47 复制带随机指针的链表 +- [ ] LC 1531 F47 寻找旋转排序数组中的最小值 +- [ ] LC 1545 F46 单词拆分 +- [ ] LC 1387 F46 二叉树的序列化与反序列化 +- [ ] LC 1222 F45 验证IP地址 +- [ ] LC 1142 F45 和为K的子数组 +- [ ] LC 1475 F44 长度最小的子数组 +- [ ] LC 1548 F42 只出现一次的数字 +- [ ] LC 1460 F41 基本计算器 +- [ ] LC 143 F41 用两个栈实现队列 +- [ ] LC 1196 F41 对角线遍历 +- [ ] LC 1230 F40 LFU缓存 +- [ ] LC 1282 F40 移掉K位数字 +- [ ] LC 111 F40 二叉搜索树与双向链表 +- [ ] LC 1907 F40 手撕归并排序 +- [ ] LC 1477 F39 课程表 +- [ ] LC 978 F39 每日温度 +- [ ] LC 1924 F38 检测循环依赖 +- [ ] LC 1903 F38 排序奇升偶降链表 +- [ ] LC 730 F38 二叉树的完全性检验 +- [ ] LC 83 F37 二叉搜索树的第k大节点 +- [ ] LC 1673 F36 盛最多水的容器 +- [ ] LC 139 F36 青蛙跳台阶问题 +- [ ] LC 1605 F36 单词搜索 +- [ ] LC 1644 F35 组合总和 II +- [ ] LC 1634 F34 Pow(x，n) +- [ ] LC 1637 F34 全排列 II +- [ ] LC 1904 F34 圆环回原点问题 +- [ ] LC 1625 F34 螺旋矩阵 II +- [ ] LC 1178 F34 零钱兑换 II +- [ ] LC 1629 F34 跳跃游戏 +- [ ] LC 1539 F34 二叉树的后序遍历 +- [ ] LC 98 F34 数组中的逆序对 +- [ ] LC 1623 F33 旋转链表 +- [ ] LC 1610 F33 搜索二维矩阵 +- [ ] LC 1658 F33 删除排序数组中的重复项 +- [ ] LC 142 F33 斐波那契数列 +- [ ] LC 1561 F32 买卖股票的最佳时机 III +- [ ] LC 112 F32 连续子数组的最大和 +- [ ] LC 115 F32 最小的k个数 +- [ ] LC 1240 F31 删除二叉搜索树中的节点 +- [ ] LC 1677 F31 整数反转 +- [ ] LC 1250 F30 字典序的第K小数字 +- [ ] LC 125 F30 调整数组顺序使奇数位于偶数前面 +- [ ] LC 1454 F30 二叉搜索树中第K小的元素 +- [ ] LC 1459 F29 用队列实现栈 +- [ ] LC 1549 F29 分发糖果 +- [ ] LC 73 F29 圆圈中最后剩下的数字 +- [ ] LC 1609 F28 颜色分类 +- [ ] LC 1355 F28 矩阵中的最长递增路径 +- [ ] LC 140 F27 二维数组中的查找 +- [ ] LC 1356 F27 奇偶链表 +- [ ] LC 1593 F27 解码方法 +- [ ] LC 1130 F26 另一个树的子树 +- [ ] LC 1300 F26 打乱数组 +- [ ] LC 1559 F26 验证回文串 +- [ ] LC 129 F25 树的子结构 +- [ ] LC 1476 F25 实现 Trie (前缀树) +- [ ] LC 1675 F25 回文数 +- [ ] LC 1495 F25 轮转数组 +- [ ] LC 1639 F25 跳跃游戏 II +- [ ] LC 1245 F25 两数相加 II +- [ ] LC 128 F25 二叉树的镜像 +- [ ] LC 1397 F24 寻找重复数 +- [ ] LC 1389 F24 数据流的中位数 +- [ ] LC 1570 F24 二叉树展开为链表 +- [ ] LC 1588 F23 不同的二叉搜索树 +- [ ] LC 1668 F23 最接近的三数之和 +- [ ] LC 1471 F23 打家劫舍 II +- [ ] LC 1564 F22 三角形最小路径和 +- [ ] LC 123 F22 顺时针打印矩阵 +- [ ] LC 1674 F22 正则表达式匹配 +- [ ] LC 1578 F22 从中序与后序遍历序列构造二叉树 +- [ ] LC 1335 F21 两个数组的交集 +- [ ] LC 1284 F21 第N个数字 +- [ ] LC 75 F20 扑克牌中的顺子 +- [ ] LC 801 F20 鸡蛋掉落 +- [ ] LC 167 F20 链表求和 +- [ ] LC 1028 F20 有效的括号字符串 +- [ ] LC 90 F20 两个链表的第一个公共节点 +- [ ] LC 1337 F20 前 K 个高频元素 +- [ ] LC 1911 F20 36进制加法 +- [ ] LC 1093 F19 有效三角形的个数 +- [ ] LC 106 F19 数组中出现次数超过一半的数字 +- [ ] LC 1516 F19 Excel表列名称 +- [ ] LC 1033 F19 最长递增子序列的个数 +- [ ] LC 1613 F19 简化路径 +- [ ] LC 1248 F19 数组中重复的数据 +- [ ] LC 715 F19 斐波那契数 +- [ ] LC 603 F19 删除字符串中的所有相邻重复项 +- [ ] LC 1916 F18 阿拉伯数字转中文数字 +- [ ] LC 1550 F18 加油站 +- [ ] LC 683 F18 最大连续1的个数 III +- [ ] LC 99 F18 二叉树中和为某一值的路径 +- [ ] LC 87 F18 在排序数组中查找数字 I +- [ ] LC 1598 F18 分隔链表 +- [ ] LC 1027 F18 24 点游戏 +- [ ] LC 1231 F17 重复的子字符串 +- [ ] LC 1573 F17 二叉树的最小深度 +- [ ] LC 1144 F17 下一个更大元素 III +- [ ] LC 1640 F17 通配符匹配 +- [ ] LC 1405 F17 完全平方数 +- [ ] LC 1621 F17 不同路径 II +- [ ] LC 1647 F17 解数独 +- [ ] LC 1340 F17 反转字符串 +- [ ] LC 1143 F17 反转字符串中的单词 III +- [ ] LC 1036 F17 最大交换 +- [ ] LC 138 F17 旋转数组的最小数字 +- [ ] LC 1247 F17 压缩字符串 +- [ ] LC 1493 F17 位1的个数 +- [ ] LC 93 F17 把数组排成最小的数 +- [ ] LC 1666 F16 四数之和 +- [ ] LC 1424 F16 只出现一次的数字 III +- [ ] LC 1587 F16 交错字符串 +- [ ] LC 17 F16 最大子矩阵 +- [ ] LC 1599 F16 最大矩形 +- [ ] LC 1474 F16 课程表 II +- [ ] LC 1191 F16 下一个更大元素 II +- [ ] LC 1442 F16 有效的字母异位词 +- [ ] LC 1584 F16 相同的树 +- [ ] LC 1420 F15 丑数 II +- [ ] LC 1600 F15 柱状图中最大的矩形 +- [ ] LC 1199 F15 目标和 +- [ ] LC 1517 F15 两数之和 II - 输入有序数组 +- [ ] LC 101 F15 二叉搜索树的后序遍历序列 +- [ ] LC 1569 F15 不同的子序列 +- [ ] LC 1633 F15 N皇后 +- [ ] LC 1416 F15 缺失数字 +- [ ] LC 1180 F15 最长回文子序列 +- [ ] LC 84 F15 最长不含重复字符的子字符串 +- [ ] LC 141 F14 数组中重复的数字 +- [ ] LC 1289 F14 至少有K个重复字符的最长子串 +- [ ] LC 1146 F14 砖墙 +- [ ] LC 1530 F14 寻找旋转排序数组中的最小值 II +- [ ] LC 1431 F14 会议室 II +- [ ] LC 1341 F14 整数拆分 +- [ ] LC 102 F14 从上到下打印二叉树 III +- [ ] LC 832 F14 和至少为 K 的最短子数组 +- [ ] LC 816 F13 链表的中间结点 +- [ ] LC 1427 F13 二叉树的所有路径 +- [ ] LC 949 F13 划分字母区间 +- [ ] LC 1667 F13 电话号码的字母组合 +- [ ] LC 831 F13 二叉树中所有距离为 K 的结点 +- [ ] LC 1678 F13 Z 字形变换 +- [ ] LC 710 F13 有序数组的平方 +- [ ] LC 1274 F13 分割数组的最大值 +- [ ] LC 118 F13 反转链表 +- [ ] LC 1462 F13 完全二叉树的节点个数 +- [ ] LC 1925 F13 IP地址与整数的转换 +- [ ] LC 547 F13 山脉数组中查找目标值 +- [ ] LC 1304 F12 常数时间插入、删除和获取随机元素 +- [ ] LC 959 F12 将二叉搜索树转化为排序的双向链表 +- [ ] LC 1059 F12 回文子串 +- [ ] LC 864 F12 设计循环队列 +- [ ] LC 1607 F12 组合 +- [ ] LC 1672 F12 整数转罗马数字 +- [ ] LC 900 F12 设计哈希映射 +- [ ] LC 1135 F12 字符串的排列 +- [ ] LC 1347 F12 打家劫舍 III +- [ ] LC 108 F12 字符串的排列 +- [ ] LC 91 F12 把数字翻译成字符串 +- [ ] LC 1496 F12 买卖股票的最佳时机 IV +- [ ] LC 1576 F12 将有序数组转换为二叉搜索树 +- [ ] LC 152 F12 八皇后 +- [ ] LC 1554 F12 被围绕的区域 +- [ ] LC 1513 F12 Excel表列序号 +- [ ] LC 1172 F12 连续数组 +- [ ] LC 1480 F11 计数质数 +- [ ] LC 1026 F11 验证回文字符串 Ⅱ +- [ ] LC 116 F11 复杂链表的复制 +- [ ] LC 1368 F11 去除重复字母 +- [ ] LC 1369 F11 计算右侧小于当前元素的个数 +- [ ] LC 1923 F11 字符串相减 +- [ ] LC 1330 F11 俄罗斯套娃信封问题 +- [ ] LC 1277 F11 接雨水 II +- [ ] LC 1087 F11 合并二叉树 +- [ ] LC 354 F11 让字符串成为回文串的最少插入次数 +- [ ] LC 153 F11 搜索旋转数组 +- [ ] LC 133 F11 重建二叉树 +- [ ] LC 1152 F11 省份数量（原朋友圈） +- [ ] LC 1268 F11 分割等和子集 +- [ ] LC 812 F11 按权重随机选择 +- [ ] LC 1032 F11 最长连续递增序列 +- [ ] LC 1306 F11 有序矩阵中第K小的元素 +- [ ] LC 110 F11 栈的压入、弹出序列 +- [ ] LC 1577 F11 二叉树的层次遍历 II +- [ ] LC 1518 F11 分数到小数 +- [ ] LC 1383 F10 删除无效的括号 +- [ ] LC 1174 F10 连续的子数组和 +- [ ] LC 1585 F10 恢复二叉搜索树 +- [ ] LC 713 F10 和可被 K 整除的子数组 +- [ ] LC 1014 F10 前K个高频单词 +- [ ] LC 609 F10 最长重复子串 +- [ ] LC 1298 F10 字典序排数 +- [ ] LC 1253 F10 路径总和 III +- [ ] LC 1656 F10 实现 strStr() +- [ ] LC 1512 F10 阶乘后的零 +- [ ] LC 1344 F10 至多包含 K 个不同字符的最长子串 +- [ ] LC 1611 F10 矩阵置零 +- [ ] LC 1279 F10 数字转换为十六进制数 +- [ ] LC 1910 F10 计算数组的小和 +- [ ] LC 100 F9 第一个只出现一次的字符 +- [ ] LC 953 F9 N叉树的层序遍历 +- [ ] LC 1547 F9 只出现一次的数字 II +- [ ] LC 1624 F9 第k个排列 +- [ ] LC 1077 F9 三个数的最大乘积 +- [ ] LC 48 F9 汉诺塔问题 +- [ ] LC 1575 F9 有序链表转换二叉搜索树 +- [ ] LC 1553 F9 分割回文串 +- [ ] LC 964 F9 打开转盘锁 +- [ ] LC 1008 F9 划分为k个相等的子集 +- [ ] LC 1432 F9 会议室 +- [ ] LC 92 F9 和为s的连续正数序列 +- [ ] LC 1567 F9 填充每个节点的下一个右侧节点指针 II +- [ ] LC 693 F8 腐烂的橘子 +- [ ] LC 1635 F8 字母异位词分组 +- [ ] LC 742 F8 验证栈序列 +- [ ] LC 80 F8 滑动窗口的最大值 +- [ ] LC 858 F8 矩形重叠 +- [ ] LC 1482 F8 快乐数 +- [ ] LC 1534 F8 逆波兰表达式求值 +- [ ] LC 1378 F8 累加数 +- [ ] LC 817 F8 爱吃香蕉的珂珂 +- [ ] LC 701 F8 区间列表的交集 +- [ ] LC 1255 F8 无重叠区间 +- [ ] LC 776 F8 排序数组 +- [ ] LC 1604 F8 删除排序数组中的重复项 II +- [ ] LC 137 F8 矩阵中的路径 +- [ ] LC 781 F8 子数组的最小值之和 +- [ ] LC 1446 F8 除自身以外数组的乘积 +- [ ] LC 132 F8 剪绳子 +- [ ] LC 1535 F8 直线上最多的点数 +- [ ] LC 932 F8 基本计算器 III +- [ ] LC 1557 F8 单词接龙 +- [ ] LC 1451 F8 数字 1 的个数 +- [ ] LC 1566 F8 杨辉三角 +- [ ] LC 933 F8 二叉搜索树中的插入操作 +- [ ] LC 1168 F8 二叉搜索树的最小绝对差 +- [ ] LC 1238 F7 用最少数量的箭引爆气球 +- [ ] LC 1919 F7 两个有序数组第k小的数 +- [ ] LC 1252 F7 找到字符串中所有字母异位词 +- [ ] LC 107 F7 1～n整数中1出现的次数 +- [ ] LC 1671 F7 罗马数字转整数 +- [ ] LC 1915 F7 中文数字转阿拉伯数字 +- [ ] LC 982 F7 行星碰撞 +- [ ] LC 1470 F7 最短回文串 +- [ ] LC 1001 F7 最大栈 +- [ ] LC 37 F7 LRU缓存 +- [ ] LC 1234 F7 132模式 +- [ ] LC 1648 F7 有效的数独 +- [ ] LC 1350 F7 递增的三元子序列 +- [ ] LC 1525 F7 至多包含两个不同字符的最长子串 +- [ ] LC 1411 F7 整数转换英文表示 +- [ ] LC 42 F7 最小K个数 +- [ ] LC 1914 F7 二叉树的下一个节点 +- [ ] LC 799 F7 根据前序和后序遍历构造二叉树 +- [ ] LC 943 F7 N叉树的最大深度 +- [ ] LC 793 F7 最大频率栈 +- [ ] LC 1511 F7 二叉搜索树迭代器 +- [ ] LC 939 F7 重构字符串 +- [ ] LC 1453 F7 2的幂 +- [ ] LC 81 F7 不用加减乘除做加法 +- [ ] LC 1926 F7 双栈排序 +- [ ] LC 1627 F6 插入区间 +- [ ] LC 842 F6 山脉数组的峰顶索引 +- [ ] LC 667 F6 将数组分成和相等的三个部分 +- [ ] LC 72 F6 股票的最大利润 +- [ ] LC 526 F6 航班预订统计 +- [ ] LC 1842 F6 千位分隔数 +- [ ] LC 1048 F6 找到 K 个最接近的元素 +- [ ] LC 1275 F6 最长回文串 +- [ ] LC 472 F6 从链表中删去总和值为零的连续节点 +- [ ] LC 1239 F6 根据字符出现频率排序 +- [ ] LC 413 F6 删除子文件夹 +- [ ] LC 113 F6 数据流中的中位数 +- [ ] LC 369 F6 网格中的最短路径 +- [ ] LC 1121 F6 最短无序连续子数组 +- [ ] LC 1206 F6 预测赢家 +- [ ] LC 1292 F6 判断子序列 +- [ ] LC 1372 F6 戳气球 +- [ ] LC 1551 F6 克隆图 +- [ ] LC 245 F6 绝对差不超过限制的最长连续子数组 +- [ ] LC 1922 F6 立方根 +- [ ] LC 1449 F6 二叉搜索树的最近公共祖先 +- [ ] LC 1544 F6 单词拆分 II +- [ ] LC 1520 F6 最大间距 +- [ ] LC 928 F6 数据流中的第K大元素 +- [ ] LC 1159 F6 有序数组中的单一元素 +- [ ] LC 1617 F6 二进制求和 +- [ ] LC 1263 F6 数组中两个数的最大异或值 +- [ ] LC 1905 F6 求区间最小数乘区间和的最大值 +- [ ] LC 130 F6 合并两个排序的链表 +- [ ] LC 1909 F6 木头切割问题 +- [ ] LC 1729 F5 将 x 减到 0 的最小操作数 +- [ ] LC 77 F5 n个骰子的点数 +- [ ] LC 1291 F5 UTF-8 编码验证 +- [ ] LC 1565 F5 杨辉三角 II +- [ ] LC 1461 F5 矩形面积 +- [ ] LC 162 F5 递归乘法 +- [ ] LC 991 F5 原子的数量 +- [ ] LC 170 F5 旋转矩阵 +- [ ] LC 773 F5 分割数组 +- [ ] LC 1303 F5 O(1) 时间插入、删除和获取随机元素 - 允许重复 +- [ ] LC 1183 F5 找树左下角的值 +- [ ] LC 1655 F5 两数相除 +- [ ] LC 1012 F5 不同岛屿的数量 +- [ ] LC 916 F5 判断二分图 +- [ ] LC 120 F5 打印从1到最大的n位数 +- [ ] LC 915 F5 第 K 个最小的素数分数 +- [ ] LC 633 F5 两个非重叠子数组的最大和 +- [ ] LC 124 F5 二进制中1的个数 +- [ ] LC 1595 F5 格雷编码 +- [ ] LC 79 F5 队列的最大值 +- [ ] LC 920 F5 森林中的兔子 +- [ ] LC 423 F5 设计跳表 +- [ ] LC 952 F5 序列化和反序列化 N 叉树 +- [ ] LC 995 F5 删除注释 +- [ ] LC 88 F5 翻转单词顺序 +- [ ] LC 1888 F5 删掉一个元素以后全为 1 的最长子数组 +- [ ] LC 192 F5 字符串压缩 +- [ ] LC 1280 F5 左叶子之和 +- [ ] LC 1241 F5 序列化和反序列化二叉搜索树 +- [ ] LC 313 F5 最多可以参加的会议数目 +- [ ] LC 119 F5 正则表达式匹配 +- [ ] LC 1212 F5 滑动窗口中位数 +- [ ] LC 85 F5 0～n-1中缺失的数字 +- [ ] LC 1236 F4 四数相加 II +- [ ] LC 786 F4 最大为 N 的数字组合 +- [ ] LC 368 F4 学生们参加各科测试的次数 +- [ ] LC 1359 F4 和等于 k 的最长子数组长度 +- [ ] LC 1467 F4 存在重复元素 +- [ ] LC 1616 F4 文本左右对齐 +- [ ] LC 411 F4 规划兼职工作 +- [ ] LC 1160 F4 最小时间差 +- [ ] LC 892 F4 保持城市天际线 +- [ ] LC 1019 F4 最长同值路径 +- [ ] LC 1311 F4 查找和最小的K对数字 +- [ ] LC 1552 F4 分割回文串 II +- [ ] LC 1447 F4 删除链表中的节点 +- [ ] LC 1594 F4 子集 II +- [ ] LC 1443 F4 为运算表达式设计优先级 +- [ ] LC 60 F4 部分排序 +- [ ] LC 720 F4 监控二叉树 +- [ ] LC 766 F4 按奇偶排序数组 II +- [ ] LC 1649 F4 搜索插入位置 +- [ ] LC 1218 F4 火柴拼正方形 +- [ ] LC 1120 F4 杀死进程 +- [ ] LC 254 F4 对角线遍历 II +- [ ] LC 1497 F4 重复的DNA序列 +- [ ] LC 1297 F4 字符串中的第一个唯一字符 +- [ ] LC 1319 F4 水壶问题 +- [ ] LC 1913 F4 翻转URL字符串里的单词 +- [ ] LC 905 F4 旋转字符串 +- [ ] LC 1921 F4 判断一个点是否在三角形内 +- [ ] LC 511 F4 删除被覆盖区间 +- [ ] LC 1657 F4 移除元素 +- [ ] LC 1479 F4 同构字符串 +- [ ] LC 1654 F4 串联所有单词的子串 +- [ ] LC 1404 F4 摆动排序 +- [ ] LC 1363 F4 拼接最大数 +- [ ] LC 815 F4 石子游戏 +- [ ] LC 1353 F4 验证二叉树的前序序列化 +- [ ] LC 1072 F4 平方数之和 +- [ ] LC 56 F4 一次编辑 +- [ ] LC 326 F4 分裂二叉树的最大乘积 +- [ ] LC 973 F4 网络延迟时间 +- [ ] LC 934 F4 二叉搜索树中的搜索 +- [ ] LC 1360 F4 摆动排序 II +- [ ] LC 51 F4 迷路的机器人 +- [ ] LC 1481 F4 移除链表元素 +- [ ] LC 1334 F4 两个数组的交集 II +- [ ] LC 1315 F4 给单链表加一 +- [ ] LC 117 F4 删除链表的节点 +- [ ] LC 601 F3 最后一块石头的重量 II +- [ ] LC 89 F3 礼物的最大价值 +- [ ] LC 401 F3 移除无效的括号 +- [ ] LC 445 F3 反转每对括号间的子串 +- [ ] LC 1227 F3 岛屿的周长 +- [ ] LC 783 F3 按奇偶排序数组 +- [ ] LC 1743 F3 可以到达的最远建筑 +- [ ] LC 619 F3 拼写单词 +- [ ] LC 1216 F3 供暖器 +- [ ] LC 1083 F3 任务调度器 +- [ ] LC 807 F3 救生艇 +- [ ] LC 1004 F3 乘积小于K的子数组 +- [ ] LC 405 F3 飞机座位分配概率 +- [ ] LC 1003 F3 买卖股票的最佳时机含手续费 +- [ ] LC 1748 F3 到家的最少跳跃次数 +- [ ] LC 154 F3 硬币 +- [ ] LC 555 F3 树的直径 +- [ ] LC 1307 F3 组合总和 Ⅳ +- [ ] LC 984 F3 图像渲染 +- [ ] LC 849 F3 数组中的最长山脉 +- [ ] LC 1054 F3 寻找重复的子树 +- [ ] LC 1537 F3 对链表进行插入排序 +- [ ] LC 1244 F3 等差数列划分 II - 子序列 +- [ ] LC 642 F3 最长等差数列 +- [ ] LC 1073 F3 最小区间 +- [ ] LC 121 F3 数值的整数次方 +- [ ] LC 878 F3 单词的压缩编码 +- [ ] LC 1278 F3 根据身高重建队列 +- [ ] LC 721 F3 连续差相同的数字 +- [ ] LC 809 F3 在圆内随机生成点 +- [ ] LC 868 F3 连续整数求和 +- [ ] LC 1260 F3 替换后的最长重复字符 +- [ ] LC 1052 F3 最大二叉树 +- [ ] LC 1632 F3 N皇后 II +- [ ] LC 956 F3 员工空闲时间 +- [ ] LC 1618 F3 加一 +- [ ] LC 1380 F3 二维区域和检索 - 矩阵不可变 +- [ ] LC 1381 F3 区域和检索 - 数组不可变 +- [ ] LC 979 F3 单调递增的数字 +- [ ] LC 95 F3 数组中数字出现的次数 II +- [ ] LC 475 F3 单字符重复子串的最大长度 +- [ ] LC 1881 F3 检查数组对是否可以被 k 整除 +- [ ] LC 823 F3 优势洗牌 +- [ ] LC 78 F3 构建乘积数组 +- [ ] LC 846 F3 字母移位 +- [ ] LC 473 F3 掷骰子的N种方法 +- [ ] LC 714 F3 最接近原点的 K 个点 +- [ ] LC 481 F3 层数最深叶子节点的和 +- [ ] LC 188 F3 链表相交 +- [ ] LC 1912 F3 36进制减法 +- [ ] LC 670 F3 有序数组中的缺失元素 +- [ ] LC 462 F3 形成三的最大倍数 +- [ ] LC 1373 F3 稀疏矩阵的乘法 +- [ ] LC 1138 F3 寻找最近的回文数 +- [ ] LC 135 F3 机器人的运动范围 +- [ ] LC 1379 F3 岛屿数量 II +- [ ] LC 1308 F3 摆动序列 +- [ ] LC 1242 F3 找到所有数组中消失的数字 +- [ ] LC 136 F3 替换空格 +- [ ] LC 737 F3 翻转等价二叉树 +- [ ] LC 1494 F3 颠倒二进制位 +- [ ] LC 1920 F3 反转双向链表 +- [ ] LC 852 F3 将数组拆分成斐波那契序列 +- [ ] LC 1568 F3 填充每个节点的下一个右侧节点指针 +- [ ] LC 617 F3 从二叉搜索树到更大和树 +- [ ] LC 877 F3 字符的最短距离 +- [ ] LC 190 F3 移除重复节点 +- [ ] LC 1038 F3 乘法表中第k小的数 +- [ ] LC 395 F3 统计封闭岛屿的数目 +- [ ] LC 97 F3 二叉树的深度 +- [ ] LC 885 F3 二叉树剪枝 +- [ ] LC 55 F2 返回倒数第 k 个节点 +- [ ] LC 300 F2 灯泡开关 III +- [ ] LC 447 F2 树节点的第 K 个祖先 +- [ ] LC 391 F2 可被三整除的最大和 +- [ ] LC 1394 F2 单词规律 +- [ ] LC 962 F2 到达终点数字 +- [ ] LC 1343 F2 扁平化嵌套列表迭代器 +- [ ] LC 1264 F2 强密码检验器 +- [ ] LC 1415 F2 火星词典 +- [ ] LC 819 F2 最长的斐波那契子序列的长度 +- [ ] LC 1421 F2 丑数 +- [ ] LC 69 F2 二叉树的最近公共祖先 +- [ ] LC 1366 F2 最大单词长度乘积 +- [ ] LC 1473 F2 添加与搜索单词 - 数据结构设计 +- [ ] LC 1281 F2 青蛙过河 +- [ ] LC 901 F2 设计哈希集合 +- [ ] LC 624 F2 不相交的线 +- [ ] LC 564 F2 复写零 +- [ ] LC 562 F2 最短公共超序列 +- [ ] LC 406 F2 铺瓷砖 +- [ ] LC 1010 F2 计数二进制子串 +- [ ] LC 1333 F2 安卓系统手势解锁 +- [ ] LC 1258 F2 全 O(1) 的数据结构 +- [ ] LC 431 F2 尽可能使字符串相等 +- [ ] LC 1873 F2 最多 K 次交换相邻数位后得到的最小整数 +- [ ] LC 1270 F2 第三大的数 +- [ ] LC 1111 F2 有效的正方形 +- [ ] LC 673 F2 在 D 天内送达包裹的能力 +- [ ] LC 622 F2 最少交换次数来组合所有的 1 +- [ ] LC 696 F2 坏了的计算器 +- [ ] LC 28 F2 婴儿名字 +- [ ] LC 1193 F2 二叉搜索树中的众数 +- [ ] LC 545 F2 交替打印FooBar +- [ ] LC 904 F2 所有可能的路径 +- [ ] LC 1455 F2 求众数 II +- [ ] LC 1119 F2 两个字符串的删除操作 +- [ ] LC 131 F2 剪绳子 II +- [ ] LC 164 F2 三步问题 +- [ ] LC 1425 F2 较小的三数之和 +- [ ] LC 922 F2 第K个语法符号 +- [ ] LC 1271 F2 等差数列划分 +- [ ] LC 685 F2 查找常用字符 +- [ ] LC 1352 F2 重新安排行程 +- [ ] LC 1009 F2 数组的度 +- [ ] LC 1217 F2 一和零 +- [ ] LC 870 F2 最大人工岛 +- [ ] LC 525 F2 删点成林 +- [ ] LC 505 F2 颜色交替的最短路径 +- [ ] LC 1018 F2 “马”在棋盘上的概率 +- [ ] LC 1887 F2 并行课程 II +- [ ] LC 707 F2 不同路径 III +- [ ] LC 614 F2 地图分析 +- [ ] LC 1210 F2 密钥格式化 +- [ ] LC 1173 F2 通过删除字母匹配到字典里最长单词 +- [ ] LC 76 F2 把字符串转换成整数 +- [ ] LC 126 F2 表示数值的字符串 +- [ ] LC 1288 F2 旋转函数 +- [ ] LC 331 F2 包含所有三种字符的子字符串数目 +- [ ] LC 977 F2 删除与获得点数 +- [ ] LC 1207 F2 最大连续1的个数 +- [ ] LC 70 F2 二叉搜索树的最近公共祖先 +- [ ] LC 1157 F2 01 矩阵 +- [ ] LC 1301 F2 赎金信 +- [ ] LC 1339 F2 反转字符串中的元音字母 +- [ ] LC 30 F2 字母与数字 +- [ ] LC 105 F2 从上到下打印二叉树 +- [ ] LC 700 F2 二叉树的垂序遍历 +- [ ] LC 1184 F2 从始点到终点的所有路径 +- [ ] LC 74 F2 平衡二叉树 +- [ ] LC 1158 F2 反转字符串 II +- [ ] LC 902 F2 香槟塔 +- [ ] LC 1824 F2 删除最短的子数组使剩余数组有序 +- [ ] LC 1856 F2 第 k 个缺失的正整数 +- [ ] LC 548 F2 拼车 +- [ ] LC 1037 F2 修剪二叉搜索树 +- [ ] LC 1053 F2 两数之和 IV - 输入 BST +- [ ] LC 930 F2 滑动谜题 +- [ ] LC 1626 F2 最后一个单词的长度 +- [ ] LC 1185 F2 二叉搜索树中的中序后继 II +- [ ] LC 1302 F2 链表随机节点 +- [ ] LC 134 F2 从尾到头打印链表 +- [ ] LC 1293 F2 完美矩形 +- [ ] LC 1392 F2 Nim 游戏 +- [ ] LC 1374 F2 最小高度树 +- [ ] LC 971 F2 前缀和后缀搜索 +- [ ] LC 808 F2 索引处的解码字符串 +- [ ] LC 990 F2 最小窗口子序列 +- [ ] LC 103 F2 数字序列中某一位的数字 +- [ ] LC 1342 F2 4的幂 +- [ ] LC 1202 F2 递增子序列 +- [ ] LC 529 F2 按序打印 +- [ ] LC 698 F2 数组形式的整数加法 +- [ ] LC 770 F2 环形子数组的最大和 +- [ ] LC 1483 F2 数字范围按位与 +- [ ] LC 573 F2 不同字符的最小子序列 +- [ ] LC 436 F2 交替打印字符串 +- [ ] LC 1365 F2 灯泡开关 +- [ ] LC 645 F2 长度为 K 的无重复字符子串 +- [ ] LC 1375 F2 最佳买卖股票时机含冷冻期 +- [ ] LC 476 F2 一年中的第几天 +- [ ] LC 94 F2 和为s的两个数字 +- [ ] LC 847 F2 访问所有节点的最短路径 +- [ ] LC 861 F2 树中距离之和 +- [ ] LC 71 F2 求1+2+…+n +- [ ] LC 144 F2 最小差 +- [ ] LC 496 F2 最大的以 1 为边界的正方形 +- [ ] LC 33 F2 单词转换 +- [ ] LC 1169 F2 扫雷游戏 +- [ ] LC 1358 F2 3的幂 +- [ ] LC 936 F2 基本计算器 IV +- [ ] LC 1740 F2 合并两个链表 +- [ ] LC 1351 F2 最大 BST 子树 +- [ ] LC 697 F2 等式方程的可满足性 +- [ ] LC 813 F2 盈利计划 +- [ ] LC 711 F2 三角形的最大周长 +- [ ] LC 1189 F2 迷宫 II +- [ ] LC 1110 F2 最长和谐子序列 +- [ ] LC 82 F2 丑数 +- [ ] LC 104 F2 从上到下打印二叉树 II +- [ ] LC 757 F2 下降路径最小和 +- [ ] LC 993 F2 寻找数组的中心索引 +- [ ] LC 1041 F2 非递减数列 +- [ ] LC 826 F2 转置矩阵 +- [ ] LC 127 F2 对称的二叉树 +- [ ] LC 1603 F2 搜索旋转排序数组 II +- [ ] LC 1233 F2 环形数组循环 +- [ ] LC 1068 F2 二叉树的层平均值 +- [ ] LC 739 F2 给定数字能组成的最大时间 +- [ ] LC 333 F2 根据数字二进制下 1 的数目排序 +- [ ] LC 822 F2 最低加油次数 +- [ ] LC 503 F2 将每个元素替换为右侧最大元素 +- [ ] LC 96 F2 数组中数字出现的次数 +- [ ] LC 1406 F2 第一个错误的版本 +- [ ] LC 884 F2 公交路线 +- [ ] LC 266 F2 给 N x 3 网格图涂色的方案数 +- [ ] LC 1322 F2 敲击计数器 +- [ ] LC 146 F2 交点 +- [ ] LC 1464 F2 存在重复元素 III +- [ ] LC 341 F2 删除给定值的叶子节点 +- [ ] LC 86 F2 左旋转字符串 +- [ ] LC 637 F2 节点与其祖先之间的最大差值 +- [ ] LC 1286 F2 随机数索引 +- [ ] LC 439 F2 丑数 III +- [ ] LC 344 F1 跳跃游戏 IV +- [ ] LC 178 F1 合法二叉搜索树 +- [ ] LC 751 F1 重新排列日志文件 +- [ ] LC 1370 F1 二叉树的垂直遍历 +- [ ] LC 294 F1 设计一个支持增量操作的栈 +- [ ] LC 429 F1 最后一个能进入电梯的人 +- [ ] LC 50 F1 合并排序的数组 +- [ ] LC 370 F1 元素和小于等于阈值的正方形的最大边长 +- [ ] LC 535 F1 解析布尔表达式 +- [ ] LC 1521 F1 缺失的区间 +- [ ] LC 1016 F1 员工的重要性 +- [ ] LC 917 F1 字母大小写全排列 +- [ ] LC 1312 F1 超级次方 +- [ ] LC 634 F1 两地调度 +- [ ] LC 997 F1 词典中最长的单词 +- [ ] LC 20 F1 最短超串 +- [ ] LC 941 F1 N叉树的后序遍历 +- [ ] LC 998 F1 找出第 k 小的距离对 +- [ ] LC 1246 F1 序列重建 +- [ ] LC 1022 F1 冗余连接 +- [ ] LC 855 F1 相似字符串组 +- [ ] LC 762 F1 将字符串翻转到单调递增 +- [ ] LC 1163 F1 从字符串生成二叉树 +- [ ] LC 682 F1 K 次取反后最大化的数组和 +- [ ] LC 515 F1 最深叶节点的最近公共祖先 +- [ ] LC 563 F1 受标签影响的最大值 +- [ ] LC 1030 F1 实现一个魔法字典 +- [ ] LC 1265 F1 甲板上的战舰 +- [ ] LC 420 F1 哲学家进餐 +- [ ] LC 1428 F1 粉刷房子 +- [ ] LC 988 F1 我的日程安排表 I +- [ ] LC 568 F1 等差数列中缺失的数字 +- [ ] LC 1706 F1 最小不兼容性 +- [ ] LC 958 F1 设置交集大小至少为2 +- [ ] LC 456 F1 四因数 +- [ ] LC 1076 F1 K个逆序对数组 +- [ ] LC 566 F1 抛掷硬币 +- [ ] LC 421 F1 查询球队积分 +- [ ] LC 1395 F1 生命游戏 +- [ ] LC 976 F1 摘樱桃 +- [ ] LC 1408 F1 栅栏涂色 +- [ ] LC 1417 F1 回文排列 II +- [ ] LC 345 F1 时钟指针的夹角 +- [ ] LC 1386 F1 二叉树最长连续序列 +- [ ] LC 1510 F1 地下城游戏 +- [ ] LC 1713 F1 找出最具竞争力的子序列 +- [ ] LC 561 F1 二进制矩阵中的最短路径 +- [ ] LC 951 F1 扁平化多级双向链表 +- [ ] LC 516 F1 数组的相对排序 +- [ ] LC 19 F1 消失的两个数字 +- [ ] LC 1224 F1 统计重复个数 +- [ ] LC 778 F1 最小差值 II +- [ ] LC 910 F1 自定义字符串排序 +- [ ] LC 236 F1 课程安排 IV +- [ ] LC 612 F1 困于环中的机器人 +- [ ] LC 7 F1 传递信息 +- [ ] LC 184 F1 节点间通路 +- [ ] LC 671 F1 按字典序排列最小的等效字符串 +- [ ] LC 1235 F1 分发饼干 +- [ ] LC 834 F1 柠檬水找零 +- [ ] LC 1472 F1 单词搜索 II +- [ ] LC 1005 F1 两个字符串的最小ASCII删除和 +- [ ] LC 385 F1 搜索推荐系统 +- [ ] LC 1868 F1 克隆 N 叉树 +- [ ] LC 726 F1 最大宽度坡 +- [ ] LC 743 F1 使数组唯一的最小增量 +- [ ] LC 650 F1 从根到叶的二进制数之和 +- [ ] LC 719 F1 煎饼排序 +- [ ] LC 699 F1 从叶结点开始的最小字符串 +- [ ] LC 835 F1 亲密字符串 +- [ ] LC 1872 F1 在区间范围内统计奇数数目 +- [ ] LC 1814 F1 重复至少 K 次且长度为 M 的模式 +- [ ] LC 204 F1 避免洪水泛滥 +- [ ] LC 1498 F1 翻转字符串里的单词 II +- [ ] LC 523 F1 设计有限阻塞队列 +- [ ] LC 838 F1 括号的分数 +- [ ] LC 148 F1 交换数字 +- [ ] LC 1047 F1 分割数组为连续子序列 +- [ ] LC 731 F1 N 天后的牢房 +- [ ] LC 966 F1 角矩形的数量 +- [ ] LC 656 F1 飞地的数量 +- [ ] LC 186 F1 栈的最小值 +- [ ] LC 748 F1 不同的子序列 II +- [ ] LC 193 F1 URL化 +- [ ] LC 1440 F1 最短单词距离 II +- [ ] LC 1299 F1 迷你语法分析器 +- [ ] LC 2 F1 最小跳跃次数 +- [ ] LC 898 F1 使序列递增的最小交换次数 +- [ ] LC 306 F1 使网格图至少有一条有效路径的最小代价 +- [ ] LC 1056 F1 只有两个键的键盘 +- [ ] LC 574 F1 根到叶路径上的不足节点 +- [ ] LC 1228 F1 最少移动次数使数组元素相等 II +- [ ] LC 1327 F1 计算各个位数不同的数字个数 +- [ ] LC 520 F1 不相交的握手 +- [ ] LC 487 F1 二叉树着色游戏 +- [ ] LC 1254 F1 寻找右区间 +- [ ] LC 1039 F1 优美的排列 II +- [ ] LC 1050 F1 金币路径 +- [ ] LC 1797 F1 连接所有点的最小费用 +- [ ] LC 987 F1 统计不同回文子序列 +- [ ] LC 992 F1 分隔链表 +- [ ] LC 814 F1 第 N 个神奇数字 +- [ ] LC 392 F1 在受污染的二叉树中查找元素 +- [ ] LC 1310 F1 猜数字大小 +- [ ] LC 181 F1 最小高度树 +- [ ] LC 771 F1 仅仅反转字母 +- [ ] LC 299 F1 通知所有员工所需的时间 +- [ ] LC 316 F1 二叉搜索子树的最大键值和 +- [ ] LC 1508 F1 第二高的薪水 +- [ ] LC 1249 F1 排列硬币 +- [ ] LC 704 F1 最低票价 +- [ ] LC 226 F1 两个子序列的最大点积 +- [ ] LC 851 F1 猜猜这个单词 +- [ ] LC 205 F1 保证文件名唯一 +- [ ] LC 1718 F1 有序数组中差绝对值之和 +- [ ] LC 927 F1 全局倒置与局部倒置 +- [ ] LC 708 F1 在二叉树中分配硬币 +- [ ] LC 929 F1 最小化去加油站的最大距离 +- [ ] LC 1324 F1 有序转化数组 +- [ ] LC 1187 F1 完美数 +- [ ] LC 480 F1 翻转子数组得到最大的数组值 +- [ ] LC 676 F1 最小化舍入误差以满足目标 +- [ ] LC 1403 F1 锯齿迭代器 +- [ ] LC 734 F1 二倍数对数组 +- [ ] LC 733 F1 删列造序 II +- [ ] LC 728 F1 删列造序 III +- [ ] LC 426 F1 最长定差子序列 +- [ ] LC 694 F1 二叉树的堂兄弟节点 +- [ ] LC 514 F1 表现良好的最长时间段 +- [ ] LC 512 F1 有序数组中出现次数超过25%的元素 +- [ ] LC 371 F1 二进制链表转整数 +- [ ] LC 1500 F1 部门工资最高的员工 +- [ ] LC 1488 F1 删除重复的电子邮箱 +- [ ] LC 1294 F1 消除游戏 +- [ ] LC 1261 F1 从英文中重建数字 +- [ ] LC 1256 F1 字符串中的单词数 +- [ ] LC 1127 F1 分糖果 +- [ ] LC 586 F1 元素和为目标值的子矩阵数量 +- [ ] LC 1487 F1 上升的温度 +- [ ] LC 1834 F1 找出最长的超赞子字符串 +- [ ] LC 1506 F1 分数排名 +- [ ] LC 1507 F1 第N高的薪水 +- [ ] LC 1504 F1 连续出现的数字 +- [ ] LC 1503 F1 超过经理收入的员工 +- [ ] LC 544 F1 花括号展开 II +- [ ] LC 44 F1 BiNode +- [ ] LC 1099 F1 种花问题 +- [ ] LC 211 F1 一维数组的动态和 +- [ ] LC 750 F1 二叉搜索树的范围和 +- [ ] LC 1619 F1 有效数字 +- [ ] LC 909 F1 匹配子序列的单词数 +- [ ] LC 1376 F1 二维区域和检索 - 可变 +- [ ] LC 653 F1 字母切换 +- [ ] LC 465 F1 日期之间隔几天 +- [ ] LC 182 F1 栈排序 +- [ ] LC 1701 F1 石子游戏 VII +- [ ] LC 879 F1 设计链表 +- [ ] LC 1219 F1 连接词 +- [ ] LC 1295 F1 找不同 +- [ ] LC 687 F1 合并石头的最低成本 +- [ ] LC 1456 F1 汇总区间 +- [ ] LC 1040 F1 路径和 IV +- [ ] LC 1229 F1 汉明距离 +- [ ] LC 1161 F1 把二叉搜索树转换为累加树 +- [ ] LC 486 F1 快照数组 +- [ ] LC 679 F1 先序遍历构造二叉树 +- [ ] LC 758 F1 和相同的二元子数组 +- [ ] LC 195 F1 判定字符是否唯一 +- [ ] LC 1767 F1 括号的最大嵌套深度 +- [ ] LC 1589 F1 不同的二叉搜索树 II +- [ ] LC 402 F1 统计「优美子数组」 +- [ ] LC 581 F1 查找两棵二叉搜索树之和 +- [ ] LC 16 F1 T9键盘 +- [ ] LC 1721 F1 完成所有任务的最少初始能量 +- [ ] LC 54 F1 分割链表 +- [ ] LC 233 F1 重新排列句子中的单词 +- [ ] LC 1871 F1 和为奇数的子数组数目 +- [ ] LC 965 F1 IP 到 CIDR +- [ ] LC 845 F1 到最近的人的最大距离 +- [ ] LC 1029 F1 键值映射 +- [ ] LC 464 F1 验证二叉树 +- [ ] LC 309 F1 有多少小于当前数字的数字 +- [ ] LC 1646 F1 外观数列 +- [ ] LC 114 F1 包含min函数的栈 +- [ ] LC 159 F1 堆箱子 +- [ ] LC 938 F1 最多能完成排序的块 II +- [ ] LC 860 F1 图像重叠 +- [ ] LC 1145 F1 分割连接字符串 +- [ ] LC 375 F1 用户分组 +- [ ] LC 1214 F1 汉明距离总和 +- [ ] LC 538 F1 二叉树寻路 +- [ ] LC 65 F1 生存人数 +- [ ] LC 1285 F1 除法求值 +- [ ] LC 1346 F1 比特位计数 +- [ ] LC 346 F1 大小为 K 且平均值大于等于阈值的子数组数目 +- [ ] LC 1828 F1 两球之间的磁力 +- [ ] LC 1092 F1 平面上的最近距离 +- [ ] LC 540 F1 分享巧克力 +- [ ] LC 1874 F1 统计全 1 子矩形 +- [ ] LC 418 F1 可以攻击国王的皇后 +- [ ] LC 1735 F1 字符频次唯一的最小删除次数 +- [ ] LC 830 F1 黑名单中的随机数 +- [ ] LC 1466 F1 天际线问题 +- [ ] LC 414 F1 缀点成线 +- [ ] LC 1136 F1 重塑矩阵 +- [ ] LC 1917 F1 自然数数组的排序 +- [ ] LC 168 F1 删除中间节点 +- [ ] LC 36 F1 计算器 +- [ ] LC 177 F1 后继者 +- [ ] LC 684 F1 检查替换后的词是否有效 +- [ ] LC 606 F1 设计文件系统 +- [ ] LC 1203 F1 迷宫 +- [ ] LC 1385 F1 猜数字游戏 +- [ ] LC 1058 F1 单词替换 +- [ ] LC 1361 F1 无向图中连通分量的数目 +- [ ] LC 756 F1 漂亮数组 +- [ ] LC 1017 F1 三个无重叠子数组的最大和 +- [ ] LC 1558 F1 单词接龙 II +- [ ] LC 754 F1 最短的桥 +- [ ] LC 1267 F1 太平洋大西洋水流问题 +- [ ] LC 307 F1 二叉树中的列表 +- [ ] LC 228 F1 定长子串中元音的最大数目 +- [ ] LC 924 F1 在LR字符串中交换相邻字符 +- [ ] LC 380 F1 统计全为 1 的正方形子矩阵 +- [ ] LC 1232 F1 可怜的小猪 +- [ ] LC 1371 F1 超级丑数 +- [ ] LC 403 F1 交换字符使得字符串相同 +- [ ] LC 629 F1 最低成本联通所有城市 +- [ ] LC 1388 F1 最佳的碰头地点 +- [ ] LC 1060 F1 最长数对链 +- [ ] LC 800 F1 公平的糖果交换 +- [ ] LC 840 F1 相似度为 K 的字符串 +- [ ] LC 1419 F1 粉刷房子 II +- [ ] LC 1465 F1 存在重复元素 II +- [ ] LC 1468 F1 组合总和 III +- [ ] LC 981 F1 Lisp 语法解析 +- [ ] LC 483 F1 破坏回文串 +- [ ] LC 791 F1 递增顺序查找树 +- [ ] LC 690 F1 找到小镇的法官 +- [ ] LC 970 F1 使用最小花费爬楼梯 +- [ ] LC 833 F1 翻转矩阵后的得分 +- [ ] LC 267 F1 HTML 实体解析器 +- [ ] LC 252 F1 连续字符 +- [ ] LC 747 F1 有效的山脉数组 +- [ ] LC 1331 F1 贪吃蛇 +- [ ] LC 850 F1 比较含退格的字符串 +- [ ] LC 1711 F1 数组的最小偏移量 +- [ ] LC 942 F1 N叉树的前序遍历 +- [ ] LC 688 F1 可以被一步捕获的棋子数 +- [ ] LC 738 F1 按递增顺序显示卡牌 +- [ ] LC 1813 F1 乘积为正数的最长子数组长度 +- [ ] LC 1132 F1 至少有5名直接下属的经理 +- [ ] LC 918 F1 二叉搜索树节点最小距离 +- [ ] LC 1329 F1 设计推特 +- [ ] LC 225 F1 商品折扣后的最终价格 +- [ ] LC 432 F1 删除字符串中的所有相邻重复项 II +- [ ] LC 1200 F1 翻转对 +- [ ] LC 1687 F1 速算机器人 +- [ ] LC 761 F1 三等分 +- [ ] LC 1197 F1 下一个更大元素 I +- [ ] LC 452 F1 删除一次得到子数组最大和 +- [ ] LC 441 F1 最长快乐字符串 +- [ ] LC 425 F1 黄金矿工 +- [ ] LC 359 F1 跳跃游戏 III +- [ ] LC 1317 F1 有效的完全平方数 +- [ ] LC 166 F1 堆盘子 +- [ ] LC 517 F1 打印零与奇偶数 +- [ ] LC 199 F1 覆盖 +- [ ] LC 587 F1 水资源分配优化 +- [ ] LC 311 F1 电影评分 +- [ ] LC 1182 F1 自由之路 +- [ ] LC 1790 F1 两个稀疏向量的点积 +- [ ] LC 1766 F1 最大网络秩 +- [ ] LC 1063 F1 子数组最大平均数 I +- [ ] LC 1426 F1 各位相加 +- [ ] LC 1106 F1 范围求和 II +- [ ] LC 947 F1 情侣牵手 +- [ ] LC 1287 F1 整数替换 +- [ ] LC 989 F1 自除数 +- [ ] LC 528 F1 每日新用户统计 +- [ ] LC 187 F1 环路检测 +- [ ] LC 258 F1 生成数组 +- [ ] LC 1000 F1 1比特与2比特字符 +- [ ] LC 1818 F1 石子游戏 V +- [ ] LC 1114 F1 设计内存文件系统 +- [ ] LC 1323 F1 轰炸敌人 +- [ ] LC 469 F1 餐盘栈 +- [ ] LC 1860 F1 换酒问题 +- [ ] LC 1150 F1 二叉树中最长的连续序列 +- [ ] LC 1095 F1 在系统中查找重复文件 +- [ ] LC 695 F1 K 个不同整数的子数组 +- [ ] LC 34 F1 直方图的水量 +- [ ] LC 657 F1 链表中的下一个更大节点 +- [ ] LC 1133 F1 员工薪水中位数 +- [ ] LC 1377 F1 区域和检索 - 数组可修改 +- [ ] LC 703 F1 不含 AAA 或 BBB 的字符串 +- [ ] LC 1880 F1 满足条件的子序列数目 +- [ ] LC 784 F1 水果成篮 +- [ ] LC 759 F1 独特的电子邮件地址 +- [ ] LC 493 F1 矩阵区域和 +- [ ] LC 769 F1 完全二叉树插入器 +- [ ] LC 828 F1 具有所有最深结点的最小子树 +- [ ] LC 732 F1 最高的广告牌 +- [ ] LC 1407 F1 搜寻名人 +- [ ] LC 1429 F1 验证前序遍历序列二叉搜索树 +- [ ] LC 1140 F1 矩阵中最长的连续1线段 +- [ ] LC 484 F1 段式回文 +- [ ] LC 1181 F1 在每个树行中找最大值 +- [ ] LC 1069 F1 函数的独占时间 +- [ ] LC 221 F1 数组中两元素的最大乘积 +- [ ] LC 946 F1 将 N 叉树编码为二叉树 \ No newline at end of file diff --git a/README.md b/README.md index b15ea620..097fa350 100644 --- a/README.md +++ b/README.md @@ -4,7 +4,9 @@ Run `cargo run {id}` to initialize the template submission file of "question #id Run `cargo test test_{id}` to test the solution for "question #id". +## Get hot problems +Copy code from `FetchList.js` and run it in browser console of CodeTop. ## Ref