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.

题意：给一个数组，删去其中相同的元素，最后返回无重复元素的个数len，并且这个数组的前Len个元素就是无重复的这些元素。考虑使用Left，right两个指针来看。具体过程在函数中有注释，只需要注意一些边界的处理。值得注意的是，不允许使用额外的空间，不允许重开数组。

//29ms ACclass Solution {public:    int removeDuplicates(vector<int>& nums) {        int left = 0, right = left+1;        //left存放最后无重复数组的末地址        int len = nums.size();        if(len == 0)            return 0;        sort(nums.begin(),nums.end());        while(right < len)        {            if(nums[right] != nums[left])//当前元素不重复，放入目标数组            {                left++;                swap(nums[left],nums[right]);                right++;                while(right<len && nums[right]==nums[left])                    right++;            }            else            {                while(right<len && nums[right]==nums[left])        //找到第一个不同的数                    right++;                //把它交换到Left的后一个数的位置                if(right >= len)//已经超出数组范围                    break;                left++;                swap(nums[left],nums[right]);                right++;//先让right+1,不然这个right放的数字是刚才才换过来的数字,会错误!                while(right<len && nums[right]==nums[left])                    right++;            }        }        return left+1;    }};