leetcode/problems/128.longest-consecutive-sequence.md at master

题目地址

https://leetcode.com/problems/longest-consecutive-sequence/description/

题目描述

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.
Accepted
200,786
Submissions
485,346

思路

这是一道最最长连续数字序列长度的题目，官网给出的难度是hard.

符合直觉的做法是先排序，然后用一个变量记录最大值，遍历去更新最大值即可，

代码：

if (nums.length === 0) return 0;
let count = 1;
let maxCount = 1;
// 这里其实可以不需要排序，这么做只不过是为了方便理解
nums = [...new Set(nums)].sort((a, b) => a - b);
for (let i = 0; i < nums.length - 1; i++) {
  if (nums[i + 1] - nums[i] === 1) {
    count++;
  } else {
    if (count > maxCount) {
      maxCount = count;
    }
    count = 1;
  }
}
return Math.max(count, maxCount);

但是需要排序时间复杂度会上升，题目要求时间复杂度为 O(n), 那么我们其实可以不用排序去解决的。

思路就是将之前”排序之后，通过比较前后元素是否相差 1 来判断是否连续“的思路改为不排序而是直接遍历，然后在内部循环里面查找是否存在当前值的邻居元素，但是马上有一个问题，内部我们查找是否存在当前值的邻居元素的过程如果使用数组时间复杂度是 O(n), 那么总体的复杂度就是 O(n^2)，完全不可以接受。怎么办呢？

我们换个思路，用空间来换时间。比如用类似于 hashmap 这样的数据结构优化查询部分，将时间复杂度降低到 O(1), 代码见后面代码部分

关键点解析

空间换时间

代码

/*
 * @lc app=leetcode id=128 lang=javascript
 *
 * [128] Longest Consecutive Sequence
 *
 * https://leetcode.com/problems/longest-consecutive-sequence/description/
 *
 * algorithms
 * Hard (40.98%)
 * Total Accepted:    200.3K
 * Total Submissions: 484.5K
 * Testcase Example:  '[100,4,200,1,3,2]'
 *
 * 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.
 *
 *
 */
/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function(nums) {
  nums = new Set(nums);
  let max = 0;
  let y = 0;
  nums.forEach(x => {
    if (!nums.has(x - 1)) {
      y = x + 1;
      while (nums.has(y)) {
        y = y + 1;
      }
      max = Math.max(max, y - x); // y - x 就是从i开始到最后有多少连续的数字
    }
  });
  return max;
};