[leedcode 234] Palindrome Linked List
Given a singly linked list, determine if it is a palindrome.
Follow up:
Could you do it in O(n) time and O(1) space?
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public boolean isPalindrome(ListNode head) { /*判断给定的链表是不是回文的: 把要判断的链表分成两部分,将第二部分逆置,逐个与前一部分的val相比较,若不同返回false*/ if(head==null||head.next==null) return true; ListNode fast=head; ListNode slow=head; while(fast!=null&&fast.next!=null){//假如链表有奇数个元素,则slow指针在循环后指向正中的元素; //若链表有偶数个元素,则循环后slow指向n/2的元素位置 fast=fast.next.next; slow=slow.next; } ListNode list2=reverse(slow);//将链表后面逆置 while(list2!=null){ if(head.val!=list2.val) return false; head=head.next; list2=list2.next; } return true; } public ListNode reverse(ListNode node){ ListNode pre=null; ListNode p=node; ListNode rear=node; while(rear!=null){ rear=rear.next; p.next=pre; pre=p; p=rear; } return pre; } }