这道题可以分解为寻找链表中间节点+翻转链表:
先找到链表的中间节点,从中间节点开始翻转后半段链表,再从head和tail同时遍历看两段链表是否一致
以下是代码:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public boolean isPalindrome(ListNode head) { ListNode n1 = head; ListNode n2 = head; while(n1 != null && n1.next != null){ n1 = n1.next.next; n2 = n2.next; } ListNode tail = reverse(n2); while(head != null && tail != null){ if(head.val != tail.val){ return false; } head = head.next; tail = tail.next; } return true; } public ListNode reverse(ListNode head){ ListNode pre = null; ListNode next = null; while(head != null){ next = head.next; head.next = pre; pre = head; head = next; } return pre; } }