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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:
分析：从后向前逐一查找即可
代码：
public class Solution { public static void moveZeroes(int[] nums) { for(int i =nums.length-1; i >=0 ;i--){ if(nums[i] ==0){ int intTemp = 0; for(int j=i+1;j<nums.length;j++){ if(nums[j]!= 0){ nums[j-1] = nums[j]; nums[j] = 0; } } } } } }
网上高手的思路：现将非0的数前移，再将余下的空位补0
代码：
public class Solution { public static void moveZeroes(int[] nums) { int iCounter = 0; for(int i =0;i<nums.length;i++){ if(nums[i] ==0){ iCounter++; } else{ nums[i-iCounter] = nums[i]; } } for(int i = (nums.length-iCounter);i<nums.length;i++){ nums[i]=0; } } }
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原文地址：http://www.cnblogs.com/savageclc26/p/4827489.html
