solve #128

Author: aylei

src/lib.rs

@@ -126,3 +126,4 @@ mod n0124_binary_tree_maximum_path_sum;
 mod n0125_valid_palindrome;
 mod n0126_word_ladder_ii;
 mod n0127_word_ladder;
+mod n0128_longest_consecutive_sequence;

src/n0128_longest_consecutive_sequence.rs

@@ -0,0 +1,61 @@
+/**
+ * [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 {}
+
+// 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>) -> i32 {
+        let mut max = 0;
+        let nums = nums.into_iter().collect::<Vec<_>>();
+        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)
+    }
+}