問題描述：

Given a sorted array nums, 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 1:

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 returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn‘t matter what values are set beyond the returned length.

思路：

可以用一個下標 idx 代表新的數組的下標，初始值為0，遍歷給定數組，當nums[i] != nums[idx]時，更新idx並賦新值。

註意：

應當註意給出一個空數組的極限情況。

代碼：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size() < 2)
            return nums.size();
        int stride = 0;
        int idx = 0;
        
 
for(int i = 0; i < nums.size(); i ++){
            if(nums[i] == nums[idx]){
                continue;
            }else{
                nums[++idx] = nums[i];
            }
        }
        return (idx+1);
    }
};

leetcode26.Remove Duplicates from Sorted Array