[LeetCode]283. 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.

方法一：（我的）从数组末端开始遍历，遇到0就把后面的都向前挪，同时记录已经放在末端的0的个数以减小挪动的次数。

public class Solution {    public void moveZeroes(int[] nums) {        int n=nums.length;        int count=0;                for(int i=n-1; i>-1; i--){            if(nums[i]==0){                for(int j=i; j<n-1-count; j++){                    nums[j]=nums[j+1];                }                nums[n-1-count]=0;                count++;            }        }    }}

分析：太死板了。盯着“哪个元素是0”，但其实换个角度，“哪个元素不是0”，就把它存下来。这样子出发更简便。

方法二：

void moveZeroes(vector<int>& nums) {    int lastNonZeroFoundAt = 0;        for (int i = 0; i < nums.size(); i++) {        if (nums[i] != 0) {            nums[lastNonZeroFoundAt++] = nums[i];        }    }        for (int i = lastNonZeroFoundAt; i < nums.size(); i++) {        nums[i] = 0;    }}

方法三：用一个容器把非零元素按顺序存下来

void moveZeroes(vector<int>& nums) {    int n = nums.size();    // Count the zeroes    int numZeroes = 0;    for (int i = 0; i < n; i++) {        numZeroes += (nums[i] == 0);    }    // Make all the non-zero elements retain their original order.    vector<int> ans;    for (int i = 0; i < n; i++) {        if (nums[i] != 0) {            ans.push_back(nums[i]);        }    }    // Move all zeroes to the end    while (numZeroes--) {        ans.push_back(0);    }    // Combine the result    for (int i = 0; i < n; i++) {        nums[i] = ans[i];    }}