zoukankan      html  css  js  c++  java
  • 判断一个链表是不是回文链表

    第一种方法: 额外空间复杂度O(N) ,遍历链表时,将元素入栈,再次遍历时,从栈中弹出元素,比较两者的大小,就可以判断是不是回文链表
    第二种方法:利用快慢指针,先找到链表的中间位置,然后反转链表的后半部分,再分别从链表两头遍历比较大小,最后将链表恢复为原始结构

    public class PalindromeLinkedList {
    
        
        public static void main(String[] args) {
            
            Node head = new Node(1);
            head.next = new Node(3);
            head.next.next = new Node(5);
            head.next.next.next = new Node(5);
            head.next.next.next.next  = new Node(3);
            head.next.next.next.next.next  = new Node(1);
            
            display(head);
            
            //快指针每次走二步,慢指针每次走一步,循环结束后,当节点数量为奇数时,slow来到了中间位置,当节点数量为偶数时,slow来到了中间位置(虚线)的前一个
            Node fast = head;
            Node slow = head;
            while(fast.next != null && fast.next.next != null){
                 fast = fast.next.next;
                 slow = slow.next;
            }
            //这里的slow就表示中间节点,记录slow的位置
            Node middle = slow;
            
            //记录中间节点的下一个节点位置
            Node help = slow.next;
            //中间节点指向null
            slow.next = null;
            
            //从help这个节点反转链表
            Node pre = null;
            Node next = null;
            //循环结束后,pre就是最后一个节点
            while(help != null) {
                next = help.next;
                help.next = pre;
                pre = help;
                help = next;
            }
            
            //判断是否是回文
            boolean flag = true;
            Node  first = head;
            Node  second = pre;
            while(first != null && second != null) {
                if(first.value != second.value) {
                    flag = false;
                    break;
                }
               first =  first.next;
               second = second.next;
            }
            System.out.println(flag);
            
            //将链表恢复为原来的结构
            Node help_restore = pre;
            Node pre_restore = null;
            Node next_restore = null;
            while(help_restore != null) {
                next_restore = help_restore.next;
                help_restore.next = pre_restore;
                pre_restore = help_restore;
                help_restore = next_restore;
            }
            middle.next = pre_restore;
            display(head);
            
        }
    
        
        public static  void display(Node head) {
            StringBuilder sb = new StringBuilder();
            while (head != null) {
                sb.append(head.value + " -> ");
                head = head.next;
            }
            String res = sb.substring(0, sb.lastIndexOf(" -> "));
            System.out.println(res);
        }
        
        public static class Node{
            public int value; //
            public Node next; // 下一个节点
     
            public Node(int value) {
                this.value = value;
            }
        }
    }
  • 相关阅读:
    xpath教程-逐层检索和全局检索 转
    xpath教程-通过ID和Class检索 转
    minianaconda3安装
    爬取表情
    进程线程(转)
    centos 安装docker方法2
    关于Dockerfile
    根据指定规则生成游戏选项编码实战
    分布式对象存储 读书笔记
    muduo 的windows下的编译
  • 原文地址:https://www.cnblogs.com/moris5013/p/11636879.html
Copyright © 2011-2022 走看看