题目：

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:

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

代码：

不知道为什么，在本地可以运行，提交上去显示 Submission Result: Runtime Error ，感觉可能是越界？不知道。。调不出来了。。（把0和后面非0的交换）

class Solution {
public:
    vector<int> moveZeroes(vector<int>& nums) {
        for(auto zero = nums.begin(); zero < nums.end(); zero++){
            if ( *zero == 0 ){
                auto notzero = zero;
                for(notzero; notzero < nums.end(); notzero++){
                    if(*notzero != 0)
                        break;
                }  
　　　　　　　　if (zero != notzero)
                   iter_swap(zero, notzero);
            }
        }
        return nums;
    }
};

换个思路（把后面非0的元素和前面的0元素交换）：

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int last = 0, cur = 0;
        while(cur < nums.size()) {
            if(nums[cur] != 0) {
                swap(nums[last], nums[cur]);
                last++;
            }
            cur++;
        }  
    }
};

别人的代码：

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