标签：ota views length out tracking number 指针 statistic tis
Description:
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:
You must do this in-place without making a copy of the array. Minimize the total number of operations.
My Solution:
class Solution { public void moveZeroes(int[] nums) { int j = 0, len = nums.length; for(int i = 0;i < len;i++){ if(nums[i] != 0){ nums[j++] = nums[i]; } } for(int i = j;i<len;i++){ nums[i] = 0; } } }
Another Better Solution:
class Solution { public void moveZeroes(int[] nums) { int j = 0, len = nums.length; for(int i = 0;i < len;i++){ if(nums[i] != 0){ int temp = nums[j]; nums[j++] = nums[i]; nums[i] = temp; } } } }
总结:
用一个指针j表示非零元素的位置,注意和i对应的元素交换就可以了
标签：ota views length out tracking number 指针 statistic tis
原文地址：http://www.cnblogs.com/kevincong/p/7881334.html
