题解 | #判断一个链表是否为回文结构#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类 the head
     * @return bool布尔型
     */
    public boolean isPail (ListNode head) {
        // write code here
        if(head==null||head.next==null){
            return true;
        }


        ListNode slow, fast;
        ListNode dummyNode = new ListNode(-1);
        ListNode stackTop = new ListNode(-1);
        dummyNode.next = head;
        slow = dummyNode;
        fast = dummyNode;

        while (true) {
            fast = fast.next.next;
            slow = slow.next;
            ListNode newNode = new ListNode(slow.val);
            System.out.println(slow.val);
            newNode.next = stackTop;
            stackTop = newNode;
            if (fast.next == null) { // 偶数个节点情况
                break;
            } else if (fast.next.next == null) { // 奇数
                slow = slow.next;
                break;
            }
        }
        while (slow.next != null) {
            slow = slow.next;
            if(slow.val!=stackTop.val){
                return false;
            }
            stackTop = stackTop.next;
        }
        return true;
    }
}

注意的是：快指针如果在倒数第二个节点，那么进行fast.next.next操作会抛出越界异常，所以判断的条件应该是

fast.next.next!=null