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
     */
    ListNode reverse(ListNode head) {
        ListNode pre = null;
        //ListNode post = head.next;
        while(head != null) {
            ListNode post = head.next;
            head.next = pre;
            pre = head;
            head = post;
            //post = post.next;
        }
        return pre;
    }
    
    public boolean isPail (ListNode head) {
        // write code here
        if(head == null || head.next == null) {
            return true;
        }
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        slow = reverse(slow);
        while(head != null && slow != null) {
            if(head.val != slow.val) {
                return false;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
    }
}

注意：双指针的快指针循环条件为fast != null && fast.next != null