题目描述：判断一个普通的链表是否是回文链表，要求时间复杂度O(n)，空间复杂度O(1)

解决思路：

• 最简单的方法是利用栈把链表的前半段压栈，然后出栈与链表的后半段逐个比对。找中间位置的方法是快慢指针。
• 还有一种方法是利用快慢指针找到中间，然后将链表的后半部分反转，再依次进行比较，利用的是链表反转的知识。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;


public class Main_ {

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

        public ListNode(int val){
            this.val = val;
        }
    }

    //利用栈的思想
    public boolean checkPalindrom(ListNode node){
        if(node == null)
            return false;
        Stack<ListNode> stack = new Stack<>();
        ListNode fast = node;
        ListNode slow = node;
        while (fast != null && fast.next != null){
            stack.push(slow);
            slow = slow.next;
            fast = fast.next.next;
        }
        //如果为奇数个
        if(fast != null)
            slow = slow.next;
        while (!stack.isEmpty()){
            if(stack.pop().val != slow.val)
                return false;
            slow = slow.next;
        }

        return true;
    }

    //利用链表反转的思想
    public boolean checkPalindrom_(ListNode node){
        if(node == null)
            return false;
        ListNode fast = node;
        ListNode slow = node;
        while (fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }

        while (fast != null)
            slow = slow.next;
        ListNode p = slow.next;
        ListNode q = slow.next.next;
        slow.next = null;
        while(p != null){
            p.next = slow;
            slow = p;
            p = q;
            if(q.next != null)
                q = q.next;
        }
        while (slow != null){
            if(slow.val != node.val)
                return false;
            slow = slow.next;
            node = node.next;
        }

        return true;
    }

    public static void main(String[] args) {
        ListNode node1 = new Main_().new ListNode(1);
        ListNode node2 = new Main_().new ListNode(2);
        ListNode node3 = new Main_().new ListNode(3);
        ListNode node4 = new Main_().new ListNode(3);
        ListNode node5 = new Main_().new ListNode(1);
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;
        node5.next = null;
        System.out.println(new Main_().checkPalindrom(node1));
        System.out.println(new Main_().checkPalindrom(node1));
    }
}