對(duì)于一個(gè)鏈表，請(qǐng)?jiān)O(shè)計(jì)一個(gè)時(shí)間復(fù)雜度為O(n),額外空間復(fù)雜度為O(1)的算法，判斷其是否為回文結(jié)構(gòu)。

Code

//coding=utf8 /***************************************************** @Author: Alex @Created Time : Tue 03 Sep 2019 07:36:51 AM CST @File Name: palindromeList.java @Blog: https://blog.csdn.net/weixin_43336281 ****************************************************/ public class palindromeList{ public static class Node{ public int value; public Node next; public Node(int data){ this.value = data; } } public static boolean isPalindrome(Node head){ if(head == null || head.next == null) return true; Node n1 = head; Node n2 = head; while(n2.next != null && n2.next.next != null){ n1 = n1.next; n2 = n2.next.next; } n2 = n1.next; n1.next = null; Node n3 = null; while(n2 != null){ n3 = n2.next; n2.next = n1; n1 = n2; n2 = n3; } n3 = n1; //n3 is last node n2 = head; boolean res = true; while(n1 != null && n2 != null){ if(n1.value != n2.value){ res = false; break; } n1 = n1.next; n2 = n2.next; } n1 = n3.next; n3.next = null; while(n1 != null){ n2 = n1.next; n1.next = n3; n3 = n1; n1 = n2; } return res; } }

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