LeetCode 题解(214) : Move Zeroes

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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:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

题解：

三个指针，zero指向最前零的位置，start指向zero后第一个不为0的位置，end指向start之后下一个零之前最后一个不为零的位置。递增三指针。

C++版：

class Solution {
public:void moveZeroes(vector<int>& nums) {int zero = 0, start = 0, end = 0;while(true) {while(nums[zero] != 0 && zero < nums.size())zero++;if(zero == nums.size())break;start = zero + 1;while(nums[start] == 0 && start < nums.size())start++;if(start == nums.size())break;end = start;while(nums[end] != 0 && end < nums.size())end++;end--;for(int i = start; i <= end; i++) {nums[zero] = nums[i];zero++;}for(int i = zero; i <= end; i++) {nums[i] = 0;}    }}
};

Java版：

public class Solution {public void moveZeroes(int[] nums) {int zero = 0, start = 0, end = 0;while(true) {while(zero < nums.length && nums[zero] != 0)zero++;if(zero == nums.length)return;start = zero + 1;while(start < nums.length && nums[start] == 0)start++;if(start == nums.length)return;end = start;while(end < nums.length && nums[end] != 0)end++;end--;for(int i = start; i <= end; i++) {nums[zero] = nums[i];zero++;}for(int i = zero; i <= end; i++)nums[i] = 0;}}
}

Python版：

class Solution(object):def moveZeroes(self, nums):""":type nums: List[int]:rtype: void Do not return anything, modify nums in-place instead."""zero, start, end = 0, 0, 0while True:while zero < len(nums) and nums[zero] != 0:zero += 1if zero == len(nums):returnstart = zero + 1while start < len(nums) and nums[start] == 0:start += 1if start == len(nums):returnend = startwhile end < len(nums) and nums[end] != 0:end += 1end -= 1for i in range(start, end + 1):nums[zero] = nums[i]zero += 1for i in range(zero, end + 1):nums[i] = 0

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