Determine whether a linked list is a palindrome.

Title description
For a linked list, please design an algorithm with a time complexity of O(n) and an additional space complexity of O(1) to determine whether it is a palindrome.

Given the head pointer A of a linked list, please return a bool value indicating whether it is a palindrome. Ensure that the length of the linked list is less than or equal to 900.

测试样例：
1->2->2->1
返回：true

Code:

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class PalindromeList {
    
    
    public boolean chkPalindrome(ListNode head) {
    
    
        // write code here
        if (head == null){
    
    
                return false;
            }
        // write code here
        //1.找到链表的中间节点
        ListNode fast = head;
        ListNode slow = head;
        while (fast !=null && fast.next != null){
    
    
            fast =fast.next.next;
            slow = slow.next;
        }
        //2.翻转中间链表之后的节点
        ListNode cur =slow.next;//此时slow的位置就是中间节点
        while (cur != null){
    
    
            ListNode curNext = cur.next;
            cur.next = slow;
            slow = cur;
            cur = curNext;
        }
        //3.对比是否是首尾相同的回文
        while (slow != head){
    
    //当他两没遇到时
        if (slow.val != head.val){
    
    
            return false;
        }
        if (head.next == slow){
    
    //偶数情况下
            return  true;
        }
            slow = slow.next;
            head = head.next;
        }
        return true;
    }
}

result: