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.

想法：直接使用STL相關函式，先使用unique去掉重複。此時重複的元素並沒有被刪掉，只是移到了後面，該函式返回無重複元素的最後一個元素的地址

，然後在使用distance()函式得到個數

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {

        return distance(nums.begin(), unique(nums.begin(), nums.end()));
        
    }
};

另外手動實現版本

class Solution {
 
public:
    int removeDuplicates(vector<int>& nums) {
        
        if (nums.empty())
        {
            return 0;
        }
        
        int index = 0;
        
        for (int i = 1; i < nums.size(); i++ )
        {
            if ( nums[index] != nums[i] )
            {
                nums[ ++index ] = nums[i];
          
            }
        }
        return index + 1;
        
    }
};