Remove Duplicates from Sorted Array

HR，題目描述：
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.

還是不大行啊，不過這次是有預感的效率比較低，，，
實現：

int removeDuplicates(int* nums, int numsSize) {
    int length = 1;
    if(nums == NULL || (*nums == 0
 
 && numsSize == 0))
    {
        length = 0;
        return length;
    }
    for(int i = 0 ; i < numsSize-1; i++)
    {
        if(nums[i] == nums[i+1])
        {
            for(int j = i; j < numsSize-1; j++)
            {
                nums[j] = nums[j + 1];
            }
            i--;
            numsSize--;
        }
        else
 

        length++;
    }
    return length;
}

貌似關鍵就在那個for裡了，可以優化的一個方式就是去掉for，和之後的參考案例也有關，，

參考案例：

class Solution {
    public:
    int removeDuplicates(int A[], int n) {
        if(n < 2) return n;
        int id = 1;
        for(int i = 1; i < n; ++i) 
            if(A[i] != A[i-1]) A[id++] = A[i];
        return id;
    }
};

膜拜中。。。