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;
 *   }
 * }
 */
// 回文结构分析   如果是  1 2 3 2 1    1 2 1 2     12345
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 void printList(ListNode head) {
        ListNode current = head;
        while (current != null) {
            System.out.print(current.val + " -> ");
            current = current.next;
        }

    }
        
        public boolean isPail (ListNode head) {
           
            Stack<ListNode>stack1 = new Stack<>();
            ListNode p = head;
            while (p != null) {
                ListNode newNode  = new ListNode(p.val);
                stack1.push(newNode);
                p = p.next;
            }
            ListNode dummy = new ListNode(0);


            ListNode q = dummy;
            while (!stack1.isEmpty()) {
                ListNode temp = stack1.pop();
                q.next = temp;
                q = q.next;
            }
            q.next = null;
            // printList(dummy.next);


            ListNode  first = head,second = dummy.next;
            printList(first);



            while (first.val == second.val) {
                first = first.next;
                second = second.next;
                if (first == null && second == null) {
                    return true;
                } else if ((first == null && second != null) || (first != null &&
                           second == null) ) {
                    return false;
                }
            }
            return false;
        }
    }

出现问题

1.出现了一个很智障的错误，我把链表入栈，就等于把原有的链表销毁了，然后我又将原有链表和新链表比较哈哈哈

2.出栈转为链表的时候记得设置最后一个节点的next指向空

回文用入栈出栈比较好做，刚开始想到的时快慢指针，，，后面想了很久，没思路，先记下。。。