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:
解析：
自己写的代码
1 class Solution { 2 public: 3 void moveZeroes(vector<int>& nums) { 4 vector<int>::size_type i; 5 for(i=0;i < nums.size();i++) 6 { 7 vector<int>::iterator it = find(nums.begin(),nums.end()-i,0); 8 if(it == nums.end()-i) 9 break; 10 else 11 { 12 nums.erase(it); 13 nums.push_back(0); 14 } 15 } 16 } 17 };
最然通过了，但是时间长，感觉想法也比较笨，事实上通过一次遍历即可
1 class Solution { 2 public: 3 void moveZeroes(vector<int>& nums) { 4 int n = nums.size(); 5 int left = 0, right = 0; 6 while (right < n){ 7 if (nums[right]){ 8 nums[left] = nums[right]; 9 left++; 10 } 11 right++; 12 } 13 while (left < n) 14 nums[left++] = 0; 15 } 16 };
原文：http://www.cnblogs.com/raichen/p/4930095.html