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：此題比較簡單，大意是將數組中重復點刪除，然後返回新數組的長度。數組中前n個數就是新的數組。唯一難點是不能用額外空間。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() <= 1){
            return nums.size();
        }
        
 
int len = 1;                    // 記錄當前的新的數組的長度
        for (int i = 1; i < nums.size(); i++){
            if (nums[i] != nums[i - 1]){
                nums[len] = nums[i];    //將新的元素裝到前len個
                len++;
            }
        }
        return len;                     
    }
};

Leetcode 26. Remove Duplicates from Sorted Array (easy)