283. Move Zeroes

零元素归位算法

最新推荐文章于 2021-08-03 19:54:12 发布

louische

最新推荐文章于 2021-08-03 19:54:12 发布

阅读量178

点赞数

本文介绍了一种将数组中所有零元素移至末尾同时保持非零元素相对顺序的算法。该方法通过两次遍历数组实现，第一次遍历将非零元素前移，第二次将剩余位置填充为零。此外，还提供了一个更高效的解决方案，灵感源自快速排序的分区方法。

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.

This is a fairly easy problem. The only tricky part is to come up with a quick and simple solution to put all zeros to the end of the array.

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

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

in the first method, we first fill out the non zero part of the array and then tackle the part with 0s; however, in the second method, we simply exchange the position of 0 and the first non zero after this 0, which makes perfect sense.

The following solution is copied directly from Leetcode, and this answer is inspired by partition method in quick sort. However, as far as I am concerned, this solution has the flavor of the previous solution

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