Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.

void moveZeroes(int* nums, int numsSize) 
{
   int k = 0;
   for(int i = 0;i < numsSize;i++)
   {
       if(nums[i] != 0)
           nums[k++] = nums[i];
   }
   for(;k<numsSize;k++)
       nums[k] = 0;
}

2021.04.23更新
周五下班头昏脑涨，昏昏沉沉写出了如下代码，感觉自己越来越菜了：

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int nonzero = 0;
        int size = nums.size();
        for(int i = 0;i < size;i++)
        {
            //统计非0元素个数
            if(nums[i] != 0)
                nonzero += 1;
        }

        int start = 0,k = 0;
        while(k < size)
        {
            if(nums[k] != 0)
            {
                nums[start++] = nums[k];
            } 
            k++;
        }
        int i = 0;
        while(i < size - nonzero)
        {
            nums[nonzero + i++] = 0;
        }
    }
};

运行效率居然击败90%的人

Runtime: 4 ms, faster than 90.12% of C++ online submissions for Move Zeroes.
Memory Usage: 9 MB, less than 56.46% of C++ online submissions for Move Zeroes.

看了自己两年前写的代码：

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int k = 0,size = nums.size();
        for(int i = 0;i < size;i++)
        {
            if(nums[i] != 0)
                nums[k++] = nums[i];
        }
        for(;k < size;k++)
            nums[k] = 0;
    }
};

效率击败100%

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Move Zeroes.
Memory Usage: 8.9 MB, less than 83.55% of C++ online submissions for Move Zeroes.