[Algorithm] 283. Move Zeroes
Given an array
nums
, write a function to move all0
's to the end of it while maintaining the relative order of the non-zero elements.Example:
Input:[0,1,0,3,12]
Output:[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.
var moveZeroes = function(nums) {
let pos = 0;
// keep all the non-zero
for (let i = 0; i < nums.length; i++) {
if (nums[i] !== 0) {
nums[pos++] = nums[i];
}
} // add all zero numbers
for (let i = pos;i < nums.length; i++) {
nums[pos++] = 0;
}
};
This approach works the same way as above, i.e. , first fulfills one requirement and then another. The catch? It does it in a clever way. The above problem can also be stated in alternate way, " Bring all the non 0 elements to the front of array keeping their relative order same".
This is a 2 pointer approach. The fast pointer which is denoted by variable "cur" does the job of processing new elements. If the newly found element is not a 0, we record it just after the last found non-0 element. The position of last found non-0 element is denoted by the slow pointer "lastNonZeroFoundAt" variable. As we keep finding new non-0 elements, we just overwrite them at the "lastNonZeroFoundAt + 1" 'th index. This overwrite will not result in any loss of data because we already processed what was there(if it were non-0,it already is now written at it's corresponding index,or if it were 0 it will be handled later in time).
After the "cur" index reaches the end of array, we now know that all the non-0 elements have been moved to beginning of array in their original order. Now comes the time to fulfil other requirement, "Move all 0's to the end". We now simply need to fill all the indexes after the "lastNonZeroFoundAt" index with 0.
Complexity Analysis
Space Complexity : O(1)O(1). Only constant space is used.
Time Complexity: O(n). However, the total number of operations are still sub-optimal. The total operations (array writes) that code does is nn (Total number of elements).
var moveZeroes = function(nums) {
// keep all the non-zero
for (let i = 0, pos = 0; i < nums.length; i++) {
if (nums[i] !== 0) {
[nums[pos], nums[i]] = [nums[i], nums[pos]];
pos++
}
}
};
The total number of operations of the previous approach is sub-optimal. For example, the array which has all (except last) leading zeroes: [0, 0, 0, ..., 0, 1].How many write operations to the array? For the previous approach, it writes 0's n-1n−1 times, which is not necessary. We could have instead written just once. How? ..... By only fixing the non-0 element,i.e., 1.
The optimal approach is again a subtle extension of above solution. A simple realization is if the current element is non-0, its' correct position can at best be it's current position or a position earlier. If it's the latter one, the current position will be eventually occupied by a non-0 ,or a 0, which lies at a index greater than 'cur' index. We fill the current position by 0 right away,so that unlike the previous solution, we don't need to come back here in next iteration.
In other words, the code will maintain the following invariant:
All elements before the slow pointer (lastNonZeroFoundAt) are non-zeroes.
All elements between the current and slow pointer are zeroes.
Therefore, when we encounter a non-zero element, we need to swap elements pointed by current and slow pointer, then advance both pointers. If it's zero element, we just advance current pointer.
With this invariant in-place, it's easy to see that the algorithm will work.
It is a great way to kown how to maintain two pointers, one pointer 'i' which is increase by for loop, another pointer 'pos' is increased by condition, which is if(nums[i] != 0).
[Algorithm] 283. Move Zeroes的更多相关文章
- 283. Move Zeroes【easy】
283. Move Zeroes[easy] Given an array nums, write a function to move all 0's to the end of it while ...
- 283. Move Zeroes(C++)
283. Move Zeroes Given an array nums, write a function to move all 0's to the end of it while mainta ...
- LeetCode Javascript实现 283. Move Zeroes 349. Intersection of Two Arrays 237. Delete Node in a Linked List
283. Move Zeroes var moveZeroes = function(nums) { var num1=0,num2=1; while(num1!=num2){ nums.forEac ...
- 【leetcode】283. Move Zeroes
problem 283. Move Zeroes solution 先把非零元素移到数组前面,其余补零即可. class Solution { public: void moveZeroes(vect ...
- LN : leetcode 283 Move Zeroes
lc 283 Move Zeroes 283 Move Zeroes Given an array nums, write a function to move all 0's to the end ...
- 283. Move Zeroes - LeetCode
Question 283. Move Zeroes Solution 题目大意:将0移到最后 思路: 1. 数组复制 2. 不用数组复制 Java实现: 数组复制 public void moveZe ...
- 283. Move Zeroes@python
Given an array nums, write a function to move all 0's to the end of it while maintaining the relativ ...
- leetcode:283. Move Zeroes(Java)解答
转载请注明出处:z_zhaojun的博客 原文地址:http://blog.csdn.net/u012975705/article/details/50493772 题目地址:https://leet ...
- Java [Leetcode 283]Move Zeroes
题目描述: Given an array nums, write a function to move all 0's to the end of it while maintaining the r ...
随机推荐
- 2014百度之星 Party
Party Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submi ...
- Tomcat 部署 Jenkins (Linux系统)
环境说明:Linux环境,CentOS 7版本. 关于部署tomcat,见上一篇博客:https://www.cnblogs.com/lelelong/p/10252225.html 接着,在Tomc ...
- Nginx入门教程(转)
原文:https://www.cnblogs.com/qdhxhz/p/8910174.html nginx入门教程 一.概述 什么是nginx? Nginx (engine x) 是一款轻 ...
- 50道Java线程面试题分析及答案
下面是Java线程相关的热门面试题,你可以用它来好好准备面试. 1) 什么是线程?线程是操作系统能够进行运算调度的最小单位,它被包含在进程之中,是进程中的实际运作单位.程序员可以通过它进行多处理器编程 ...
- K8S学习笔记之Grafana App for Kubernetes的配置
Grafana有一套针对Kubernetes监控的APP,和Grafana-Zabbix App类似,但是配置咋一看比较麻烦,主要参数都是来自K8S. 这款APP的详细介绍请参考Grafana App ...
- .Net Core2.2 在IIS发布
.Net Core应用发布到IIS主要是如下的三个步骤: (1)在Windows Server上安装 .Net Core Hosting Bundle (2)在IIS管理器中创建IIS站点 (3)部署 ...
- mybatis映射mapper文件做like模糊查询
方法:使用concat函数连接通配符
- Spring怎么管理事务?
我们一般通过aop管理事务,就是把代码看成一个纵向有序的,然后通过aop管理事务,就好比增删改的时候需要开启一个事务,我们给他配置一个required,required就是有事务就执行事务,没有就给他 ...
- java 之 集合概述
一.集合概述 不管是哪一种数据结构,其实本质上都是容器来着,就是用来装对象的.因此,我们就要搞清楚两点:(1)如何存储(2)存储特点 1.集合 集合是 Java 中提供的一种容器,可以用来存储多个数据 ...
- 原油PETROLAEUM英语PETROLAEUM石油
petrolaeum (uncountable) Archaic spelling of petroleum petroleum See also: Petroleum Contents [hide] ...