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.

这道题让我们将一个给定数组中所有的0都移到后面，把非零数前移，要求不能改变非零数的相对应的位置关系，而且不能拷贝额外的数组，那么只能用替换法in-place来做，需要用两个指针，一个不停的向后扫，找到非零位置，然后和前面那个指针交换位置即可，参见下面的代码：

解法一：

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

下面这种解法的思路跟上面的没啥区别，写法稍稍复杂了一点。。

解法二：

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

参考资料：

https://leetcode.com/discuss/59064/c-accepted-code

https://leetcode.com/discuss/70169/1ms-java-solution

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