[LeetCode283]Move Zeroes

题目来源：https://leetcode.com/problems/move-zeroes/

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.

2个指针，将非零数放到前面，后面补零

class Solution{public:void moveZeroes(vector<int>& nums){//时间复杂度O(n)int newIndex = 0;for (int i = 0; i < nums.size(); ++i){if (nums[i] != 0){nums[newIndex++] = nums[i];}}for (; newIndex < nums.size(); ++newIndex){nums[newIndex] = 0;}}};

对数组进行一次遍历，记录下0的个数，同时将后面非零值往前移动，移动的位数为他前面的0的个数，最后根据0的个数在数组最后一次插入0,

时间复杂度为O(n)

class Solution{public:void moveZeroes(vector<int>& nums){int count = 0;for (int i = 0; i < nums.size(); ++i){if (nums[i] == 0){++count;}else{nums[i - count] = nums[i];}}for (int i = 0; i < count; ++i){nums[nums.size() - 1 - i] = 0;}}};

那么只能用替换法in-place来做，需要用两个指针，一个不停的向后扫，找到非零位置，然后和前面那个指针交换位置即可

class Solution{public:void moveZeroes(vector<int>& nums){for (int i = 0, j = 0; i < nums.size(); ++i){if (nums[i]){swap(nums[i], nums[j++]);}}}};