[LeetCode] 128. Longest Consecutive Sequence

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

Example 1:

Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

Example 2:

Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

Constraints:

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109

最长连续序列。

给定一个未排序的整数数组 nums ,找出数字连续的最长序列(不要求序列元素在原数组中连续)的长度。

请你设计并实现时间复杂度为 O(n) 的算法解决此问题。

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/longest-consecutive-sequence
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思路是在遍历数组的时候,第一次遍历先用 hashset 存住所有的元素。再遍历第二次,第二次遍历的时候,找每一个元素的 left (num - 1) 和 right (num + 1),若找到,就在 hashset 中移除,同时移动 left 和 right 的位置。最终最长的子序列长度为 right - left - 1

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int longestConsecutive(int[] nums) {
 3         // corner case
 4         if (nums == null || nums.length == 0) {
 5             return 0;
 6         }
 7 
 8         // normal case
 9         int res = 0;
10         HashSet<Integer> set = new HashSet<>();
11         for (int num : nums) {
12             set.add(num);
13         }
14         for (int i = 0; i < nums.length; i++) {
15             int down = nums[i] - 1;
16             while (set.contains(down)) {
17                 set.remove(down);
18                 down--;
19             }
20             int up = nums[i] + 1;
21             while (set.contains(up)) {
22                 set.remove(up);
23                 up++;
24             }
25             res = Math.max(res, up - down - 1);
26         }
27         return res;
28     }
29 }

 

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var longestConsecutive = function(nums) {
 6     let res = 0;
 7     let set = new Set();
 8     for (let i = 0; i < nums.length; i++) {
 9         let num = nums[i];
10         set.add(num);
11     }
12     for (let i = 0; i < nums.length; i++) {
13         let cur = nums[i];
14         let down = cur - 1;
15         while (set.has(down)) {
16             set.delete(down);
17             down--;
18         }
19         let up = cur + 1;
20         while (set.has(up)) {
21             set.delete(up);
22             up++;
23         }
24         res = Math.max(res, up - down - 1);
25     }
26     return res;
27 };

 

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posted @ 2019-10-09 11:11  CNoodle  阅读(472)  评论(0编辑  收藏  举报