[LeetCode][JavaScript]Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn‘t matter what you leave beyond the new length.

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

两个变量记录上次和上上的值，比较一下，如果出现大于等于三次了，记下下标。

删除的时候要从后往前，这样不会打乱下标的顺序。

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var removeDuplicates = function(nums) {
 6     var aRemove = [], i, previous, previous2;
 7     for(i = 0; i < nums.length; i++){
 8         if(nums[i] !== previous){
 9             //first time
10         }else if(nums[i] === previous && nums[i] !== previous2){
11             //duplicate two times
12         }else{
13             //duplicate more than two times
14             aRemove.push(i);
15         }
16         previous2 = previous;
17         previous = nums[i];
18     }
19     for(i = aRemove.length - 1; i >= 0; i--){
20         nums.splice(aRemove[i], 1);
21     }
22     return nums.length;
23 };

另一种更简洁的做法，开一个变量index，每次把正确的结果放到下标为index的位置上，index++，遍历完之后index就是目标数组的长度。

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var removeDuplicates = function(nums) {
 6     var index = 0, i, previous, previous2;
 7     for(i = 0; i < nums.length; i++){
 8         if(nums[i] !== previous){
 9             nums[index++] = nums[i];
10         }else if(nums[i] === previous && nums[i] !== previous2){
11             nums[index++] = nums[i];
12         }
13         previous2 = previous;
14         previous = nums[i];
15     }
16     return index;
17 };