题解 | Java-快慢指针-#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 class Solution {
    /**
     * 
     * @param head ListNode类 the head
     * @return bool布尔型
     */
    public boolean isPail (ListNode head) {
        // write code here
        if(head == null || head.next == null) return true;
        ListNode slow = head;
        ListNode fast = head.next;
        LinkedList<Integer> stack = new LinkedList<>();
        while(fast != null && fast.next != null) {
            stack.push(slow.val);
            slow = slow.next;
            fast = fast.next.next;
        }
        if (fast == null) { // 奇数
        } else if (fast.next == null) { // 偶数
            stack.push(slow.val);
        }
        slow = slow.next;
        while(slow != null) {
            if (slow.val != stack.pop()) return false;
            slow = slow.next;
        }
        return true;
    }
}