给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。

示例:

输入: [0,1,0,3,12]
输出: [1,3,12,0,0]

void moveZeroes(vector<int>& nums) {int n = nums.size();// Count the zeroesint numZeroes = 0;for (int i = 0; i < n; i++) {numZeroes += (nums[i] == 0);}// Make all the non-zero elements retain their original order.vector<int> ans;for (int i = 0; i < n; i++) {if (nums[i] != 0) {ans.push_back(nums[i]);}}// Move all zeroes to the endwhile (numZeroes--) {ans.push_back(0);}// Combine the resultfor (int i = 0; i < n; i++) {nums[i] = ans[i];}
}void moveZeroes(vector<int>& nums) {int lastNonZeroFoundAt = 0;// If the current element is not 0, then we need to// append it just in front of last non 0 element we found. for (int i = 0; i < nums.size(); i++) {if (nums[i] != 0) {nums[lastNonZeroFoundAt++] = nums[i];}}// After we have finished processing new elements,// all the non-zero elements are already at beginning of array.// We just need to fill remaining array with 0's.for (int i = lastNonZeroFoundAt; i < nums.size(); i++) {nums[i] = 0;}
}