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].

tips:英语读题有点失误，一开始以为非零的数字还要按从小到大排序（这种情况的时候我用快排去做了）。其实只要非零的数字按原来的顺序就可以

我的解法：冒泡排序 复杂度O(n^2)

(有点垃圾，因为最后提交发现审错题了心态有点炸，随便想了个排序)，把0当气泡不断往上排，bouble算法。

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

Top solution 时间复杂度O(n)

思想：定义一个索引insertPos = 0，遍历一次数组，遇到非零的数就执行nums[insertPos++] = nums[i]; 遍历完之后，数组insertPos之后的全填0

public void moveZeroes(int[] nums) {    if (nums == null || nums.length == 0) return;            int insertPos = 0;    for (int num: nums) {        if (num != 0) nums[insertPos++] = num;    }            while (insertPos < nums.length) {        nums[insertPos++] = 0;    }}