|
| 1 | +/** |
| 2 | + * [23] 合并 K 个升序链表 |
| 3 | + * |
| 4 | + * 给你一个链表数组,每个链表都已经按升序排列。 |
| 5 | + * 请你将所有链表合并到一个升序链表中,返回合并后的链表。 |
| 6 | + * |
| 7 | + * 示例 1: |
| 8 | + * 输入:lists = [[1,4,5],[1,3,4],[2,6]] |
| 9 | + * 输出:[1,1,2,3,4,4,5,6] |
| 10 | + * 解释:链表数组如下: |
| 11 | + * [ |
| 12 | + * 1->4->5, |
| 13 | + * 1->3->4, |
| 14 | + * 2->6 |
| 15 | + * ] |
| 16 | + * 将它们合并到一个有序链表中得到。 |
| 17 | + * 1->1->2->3->4->4->5->6 |
| 18 | + * |
| 19 | + * 示例 2: |
| 20 | + * 输入:lists = [] |
| 21 | + * 输出:[] |
| 22 | + * |
| 23 | + * 示例 3: |
| 24 | + * 输入:lists = [[]] |
| 25 | + * 输出:[] |
| 26 | + * |
| 27 | + * |
| 28 | + * 提示: |
| 29 | + * |
| 30 | + * k == lists.length |
| 31 | + * 0 <= k <= 10^4 |
| 32 | + * 0 <= lists[i].length <= 500 |
| 33 | + * -10^4 <= lists[i][j] <= 10^4 |
| 34 | + * lists[i] 按 升序 排列 |
| 35 | + * lists[i].length 的总和不超过 10^4 |
| 36 | + * |
| 37 | + */ |
| 38 | +pub struct Solution {} |
| 39 | + |
| 40 | +use crate::util::linked_list::{to_list, ListNode}; |
| 41 | + |
| 42 | +// problem: https://leetcode.cn/problems/merge-k-sorted-lists/ |
| 43 | +// discuss: https://leetcode.cn/problems/merge-k-sorted-lists/discuss/?currentPage=1&orderBy=most_votes&query= |
| 44 | + |
| 45 | +// submission codes start here |
| 46 | + |
| 47 | +use std::collections::BinaryHeap; |
| 48 | + |
| 49 | +impl PartialOrd for Box<ListNode> { |
| 50 | + fn partial_cmp(&self, other: &Box<ListNode>) -> Option<std::cmp::Ordering> { |
| 51 | + if self.val < other.val { |
| 52 | + Some(std::cmp::Ordering::Greater) |
| 53 | + } else if self.val == other.val { |
| 54 | + Some(std::cmp::Ordering::Equal) |
| 55 | + } else { |
| 56 | + Some(std::cmp::Ordering::Less) |
| 57 | + } |
| 58 | + } |
| 59 | +} |
| 60 | + |
| 61 | +impl Ord for Box<ListNode> { |
| 62 | + fn cmp(&self, other: &Self) -> std::cmp::Ordering { |
| 63 | + if self.val < other.val { |
| 64 | + std::cmp::Ordering::Greater |
| 65 | + } else if self.val == other.val { |
| 66 | + std::cmp::Ordering::Equal |
| 67 | + } else { |
| 68 | + std::cmp::Ordering::Less |
| 69 | + } |
| 70 | + } |
| 71 | +} |
| 72 | + |
| 73 | +// Definition for singly-linked list. |
| 74 | +// #[derive(PartialEq, Eq, Clone, Debug)] |
| 75 | +// pub struct ListNode { |
| 76 | +// pub val: i32, |
| 77 | +// pub next: Option<Box<ListNode>> |
| 78 | +// } |
| 79 | +// |
| 80 | +// impl ListNode { |
| 81 | +// #[inline] |
| 82 | +// fn new(val: i32) -> Self { |
| 83 | +// ListNode { |
| 84 | +// next: None, |
| 85 | +// val |
| 86 | +// } |
| 87 | +// } |
| 88 | +// } |
| 89 | +impl Solution { |
| 90 | + pub fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> { |
| 91 | + let mut heap = BinaryHeap::new(); |
| 92 | + for p in lists { |
| 93 | + if let Some(e) = p { |
| 94 | + heap.push(e) |
| 95 | + } |
| 96 | + } |
| 97 | + let mut dummy_head = Some(Box::new(ListNode { next: None, val: 0 })); |
| 98 | + let mut curr = &mut dummy_head; |
| 99 | + loop { |
| 100 | + let mut p = heap.pop(); |
| 101 | + if p.is_none() { |
| 102 | + break; |
| 103 | + } |
| 104 | + let mut p = p.unwrap(); |
| 105 | + |
| 106 | + let next = p.next.take(); |
| 107 | + if let Some(next) = next { |
| 108 | + heap.push(next); |
| 109 | + } |
| 110 | + curr.as_mut().unwrap().next = Some(p); |
| 111 | + curr = &mut curr.as_mut().unwrap().next; |
| 112 | + } |
| 113 | + dummy_head.unwrap().next |
| 114 | + } |
| 115 | +} |
| 116 | + |
| 117 | +// submission codes end |
| 118 | + |
| 119 | +#[cfg(test)] |
| 120 | +mod tests { |
| 121 | + use super::*; |
| 122 | + |
| 123 | + #[test] |
| 124 | + fn test_23() { |
| 125 | + assert_eq!( |
| 126 | + Solution::merge_k_lists(vec![ |
| 127 | + to_list(vec![1, 4, 5]), |
| 128 | + to_list(vec![1, 3, 4]), |
| 129 | + to_list(vec![2, 6]), |
| 130 | + ],), |
| 131 | + to_list(vec![1, 1, 2, 3, 4, 4, 5, 6]) |
| 132 | + ); |
| 133 | + assert_eq!(Solution::merge_k_lists(vec![]), None); |
| 134 | + assert_eq!(Solution::merge_k_lists(vec![None]), None); |
| 135 | + } |
| 136 | +} |
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