Leetcode 26. Remove Duplicates from Sorted Array

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

思路：
1. 把重复的数交换到后面去。维持一个长度逐渐增加的起始位置变化的多余的elements，例如：

1,1,1,2,2,3,4,4

当遍历到第二个１，发现等于当前最大值(=1)，则需要用left指针指向这个位置，然后用right指针从这里继续遍历；当right指针遇到２，则更新最大值等于２，且让right指向的数和left指向的数交换，left往后移动一下，right继续移动；当right指向的值还＝２，等于最大值，不需要移动left；
2. two pointer：这里是两个指针都从左边移动，昨天遇到的情况是一个左边，一个右边，都往中间移动！形式不同，但都是two pointer

//two pointerclass Solution {public:    int removeDuplicates(vector<int>& nums) {        //        int left=0,right=0;        int mx=INT_MIN;        for(int right=0;right<nums.size();right++){            if(nums[right]>mx){                 mx=nums[right];                swap(nums[left],nums[right]);                left++;            }         }        return left;    }};