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 ListNode copy(ListNode head) {
        if (head == null) return null;

        // 创建一个新的头节点
        ListNode newHead = new ListNode(head.val);
        ListNode current = newHead;
        ListNode originalCurrent = head.next;

        // 复制原链表的每个节点，并链接到新链表上
        while (originalCurrent != null) {
            current.next = new ListNode(originalCurrent.val);
            current = current.next;
            originalCurrent = originalCurrent.next;
        }

        return newHead;
    }
    public boolean isPail (ListNode head) {
        // 法一：翻转链表后对比两个链表是否相等
        if (head == null || head.next == null) {
            return true;
        }
        ListNode head1 = copy(head);
        ListNode pre = null;
        ListNode cur = head;
        ListNode res = cur;
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        while (pre != null && head1 != null) {
            if (pre.val != head1.val) {
                return false;
            }
            head1 = head1.next;
            pre = pre.next;
        }
        return pre == null && head1 == null;
    }
}

链表翻转后逐个元素判断是否相等