reverseLinkedList(翻转链表)
ReverseLinkedList(翻转链表)
链表是一种物理存储单元上非连续、非顺序的存储结构,数据元素的逻辑顺序是通过链表中的指针链接次序实现的。非连续、非顺序指的是,通过指针把一组零散的内存块串联在一起,其中每一个内存块叫做链表的节点,所以每个节点包含两部分一个data(你存放的数据),一个next(指向下一个节点)当然这里使用单向链表举例。双向链表则有两个指向节点一个next(下一个节点)一个prev(上一个节点),对比:双向链表虽然更麻烦,但是比单向链表更受欢迎,因为,他记录了上一个节点,节省了操作数据时候的时间,但是需要用内存换时间。
单向链表:

双向链表;

题目一:
给定一组数据,从尾到头进行翻转
Input: 1 ->2->3->4->5
Output: 5->5->3->2->1
思路:使用是三个变量进行控制(prev、current、next),模拟一个窗口,进行对数据的推进,进而翻转,不断改变三个元素的位置。
/**
* @author : lizi
* @date : 2021-03-02 17:05
**/
public class ListNode {
int val = 0;
ListNode next; ListNode(int x) {
val = x;
} public void setVal(int val) {
this.val = val;
} public void setNext(ListNode next) {
this.next = next;
}
}
private static ListNode reverseList(ListNode head) {
if (head == null) return null;
ListNode prev=head;
ListNode current=head.next;
// because the last Node point into null,so let fist node.next point into null
prev.next=null;
while (current!=null){
ListNode next=current.next;
// that is place where change node. change position of current and prev
current.next=prev;
prev=current;
// change current into next (in the way,the window is pushed)
current=next;
}
return prev;
}
}

题目二:
obviously,we should change position of m and n
Input:1->2->3->4->5->null ,m=2,n=5
Output:1->4>3>2>5>null
tip: actually, many companies like to make this question as ultimate question,also the question was Amazon’s question for interviewee,however, we remain can use method in reversing whole linkedList. i mean that method of pushing window.
private static ListNode reverseBetween(ListNode head, int m, int n) {
// if you can not find head, you can make the node.next to find head, todo actually, it is a node that prevent you can not find first node
ListNode dummy = new ListNode(-1);
dummy.next=head;
head=dummy;
// to find a place where you began to control list
for (int i = 1; i <m ; i++) {
head=head.next;
}
ListNode prevM=head;
ListNode mNode=prevM.next;
ListNode nNode=mNode;
ListNode postN=nNode.next;
//todo visibly, the portion is method of window that i mention in method of reversing the whole window
for (int i = m; i <n ; i++) {
ListNode next=postN.next;
postN.next=nNode;
nNode=postN;
postN=next;
}
prevM.next=nNode;
mNode.next=postN;
// you can get dummy node.next in this way ,find the first node
return dummy.next;
}
finally: to do examine
public static void main(String[] args) {
ListNode listNode1 = generateDate();
System.out.println(listNode1.toString());
// reverse whole list
ListNode reverseBetween = reverseList(generateDate());
// reverse m and n in list
ListNode listNode = reverseBetween(listNode1, 2, 4);
printForSomething(reverseBetween);
}
// just to print result
public static void printForSomething(ListNode listNode) {
if (listNode != null) {
System.out.println(listNode.val);
printForSomething(listNode.next);
}
}
the picture is what i visualize method how to run:

To summarize:
as a beginner,personally, it is abstract to imagine how the cell of linkedlist connect with each other,but it is a effective to draw what you imagine. i still want everyone to criticize what i wrong in every article.

reverseLinkedList(翻转链表)的更多相关文章
- [LeetCode] Reverse Nodes in k-Group 每k个一组翻转链表
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If ...
- C语言递归,非递归实现翻转链表
翻转链表作为,链表的常用操作,也是面试常遇到的. 分析非递归分析: 非递归用的小技巧比较多,很容易出错. 递归分析比较简单,在代码里面 代码: #include<stdio.h> #inc ...
- [LintCode] Reverse Nodes in k-Group 每k个一组翻转链表
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If ...
- lintcode 中等题: reverse linked list II 翻转链表II
题目 翻转链表 II 翻转链表中第m个节点到第n个节点的部分 样例 给出链表1->2->3->4->5->null, m = 2 和n = 4,返回1->4-> ...
- lintcode: 翻转链表
题目: 翻转链表 翻转一个链表 样例 给出一个链表1->2->3->null,这个翻转后的链表为3->2->1->null 挑战 在原地一次翻转完成 解题: 递归还 ...
- 025k个一组翻转链表
#include "000库函数.h" struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), n ...
- 【LeetCode题解】25_k个一组翻转链表(Reverse-Nodes-in-k-Group)
目录 描述 解法一:迭代 思路 Java 实现 Python 实现 复杂度分析 解法二:递归(不满足空间复杂度) 思路 Java 实现 Python 实现 复杂度分析 更多 LeetCode 题解笔记 ...
- LeetCode(15): 每k个一组翻转链表
hard! 题目描述: 给出一个链表,每 k 个节点为一组进行翻转,并返回翻转后的链表. k 是一个正整数,它的值小于或等于链表的长度.如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序. ...
- Leetcode题库——25.k个一组翻转链表
@author: ZZQ @software: PyCharm @file: ReverseList.py @time: 2018/11/6 15:13 题目要求:给出一个链表,每 k 个节点一组进行 ...
随机推荐
- 解决springmvc使用@ResponseBody返回String类型字符串中文乱码问题
问题分析: 首先: 确定的是只有当返回值是 String时才会出现中文乱码,而当返回值是Map<String, Object>或者是其它类型时,并没有中文乱码的出现. 然后找原因: 原因是 ...
- 最新 uni-app 免费教程
最新 uni-app 免费教程 uni-app 快速入门 steps 建议第一步,看完uni-app官网的首页介绍. 建议第二步,通过快速上手,亲身体验下uni-app. 建议第三步,看完<un ...
- macOS & timer & stop watch
macOS & timer & stop watch https://matthewpalmer.net/blog/2018/09/28/top-free-countdown-time ...
- Dart All In One
Dart All In One dart & flutter https://github.com/dart-lang https://github.com/dart-lang/sdk win ...
- git in depth
git in depth git delete remote branch # Deleting remote branches in Git $ git push origin --delete f ...
- qt QTimer 计时器
#include <QtCore> #include <QTimer> QTimer *timer; timer = new QTimer(this); connect(tim ...
- 大小厂必问Java后端面试题(含答案)
你好,我是yes. 这个系列的文章不会是背诵版,不是那种贴上标准答案,到时候照着答就行的面试题汇总. 我会用大白话尽量用解释性.理解性的语言来回答,但是肯定没有比平时通过一篇文章来讲解清晰,不过我尽量 ...
- GET和POST两者的区别
GET 和 POST 是 HTTP 协议中的两种发送请求的基本方法,对于前端开发者而言,几乎每天都在使用它们,再熟悉不过了,一般也都能说出几点两者的区别. 如果面试中被问到这个问题,先回答一下几点,肯 ...
- Vue学习笔记-基于CDN引入方式简单前后端分离项目学习(Vue+Element+Axios)
一 使用环境 开发系统: windows 后端IDE: PyCharm 前端IDE: VSCode 数据库: msyql,navicat 编程语言: python3.7 (Windows x86- ...
- 基于Hi3559AV100的SVP(NNIE)开发整体流程
在之后的hi3559AV100板载开发中,除了走通V4L2->VDEC->VPSS->VO(HDMI)输出,还有需要进行神经网络的开发学习,进行如face detection的开发等 ...