Update 128.longest-consecutive-sequence.md

Author: azl397985856

problems/128.longest-consecutive-sequence.md

@@ -53,7 +53,7 @@ return Math.max(count, maxCount);
 
 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),
+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`
@@ -65,48 +65,20 @@ return Math.max(count, maxCount);
 ## 代码
 
 ```js
-/*
- * @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);
+  set = new Set(nums);
   let max = 0;
-  let y = 0;
-  nums.forEach(x => {
+  let temp = 0;
+  set.forEach(x => {
     // 说明x是连续序列的开头元素
-    if (!nums.has(x - 1)) {
-      y = x + 1;
-      while (nums.has(y)) {
-        y = y + 1;
+    if (!set.has(x - 1)) {
+      temp = x + 1;
+      while (set.has(y)) {
+        temp = temp + 1;
       }
       max = Math.max(max, y - x); // y - x 就是从x开始到最后有多少连续的数字
     }