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 by modifying the input array in-place with O(1) extra memory.

Example:

Given 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：此题比较简单，大意是将数组中重复点删除，然后返回新数组的长度。数组中前n个数就是新的数组。唯一难点是不能用额外空间。

class Solution {

public:

    int removeDuplicates(vector<int>& nums) {

        if (nums.size() <= ){

            return nums.size();

        }

        int len = ;                    // 记录当前的新的数组的长度

        for (int i = ; i < nums.size(); i++){

            if (nums[i] != nums[i - ]){

                nums[len] = nums[i];    //将新的元素装到前len个

                len++;

            }

        }

        return len;

    }

};