题解 | #判断一个链表是否为回文结构#

1. 快慢指针找到一半
2. 链表逆转
3. 判断两链表是否相等

可以记忆一下快慢指针的板子，以head开头，slow.next就是以一半无论奇偶

import java.util.*;

public class Solution {

    public boolean isPail (ListNode head) {
        // 快慢指针找到一半
        if(head == null)return false;
        ListNode fast,slow,pre,p;
        fast = slow = head;
        while(fast.next!=null&&fast.next.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        
        //slow.next就是一半，然后链表逆转
        slow = slow.next;
        fast = slow;
        pre = null;
        while(fast!=null){
            ListNode t = fast.next;
            fast.next = pre;
            pre = fast;
            fast = t;
        }
        
        //判断两链表是否相等
        while(pre!=null){
            if(pre.val != head.val)return false;
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}