题解 | #判断一个链表是否为回文结构# O(1)O(n)

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) {
        // 快慢指针分成两段
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null) {
            slow = slow.next;
            fast = fast.next;
            if (fast != null) {
                fast = fast.next;
            }
        }

        // 反转后半段
        ListNode pre = null;
        ListNode current = slow;
        while (current != null) {
            ListNode next = current.next;
            current.next = pre;
            pre = current;
            current = next;
        }

        // 比对两个链表
        ListNode pointer1 = pre;
        ListNode pointer2 = head;
        while (pointer1 != null) {
            if (pointer1.val != pointer2.val) {
                return false;
            }
            pointer1 = pointer1.next;
            pointer2 = pointer2.next;
        }
        return true;
    }
}