[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/

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删除数组中重复出现三次或三次以上的数。

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。

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    var aRemove = [], i, previous, previous2;
    for(i = 0; i < nums.length; i++){
        if(nums[i] !== previous){
            //first time
        }else if(nums[i] === previous && nums[i] !== previous2){
            //duplicate two times
        }else{
            //duplicate more than two times
            aRemove.push(i);
        }
        previous2 = previous;
        previous = nums[i];
    }
    for(i = aRemove.length - 1; i >= 0; i--){
        nums.splice(aRemove[i], 1);
    }
    return nums.length;
};

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。

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    var index = 0, i, previous, previous2;
    for(i = 0; i < nums.length; i++){
        if(nums[i] !== previous){
            nums[index++] = nums[i];
        }else if(nums[i] === previous && nums[i] !== previous2){
            nums[index++] = nums[i];
        }
        previous2 = previous;
        previous = nums[i];
    }
    return index;
};