回文链表

问题描述：

        给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

示例 1：

输入：head = [1,2,2,1]
输出：true

示例 2：

输入：head = [1,2]
输出：false

解题思路：

        将链表的值复制到数组列表中，再使用双指针法判断。

//提交版


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> list = new ArrayList<>();
        ListNode curr = head;
        while (curr != null){
            list.add(curr.val);
            curr = curr.next;
        }
        int a = 0;
        int b = list.size()-1;
        while (a<b){
            if (list.get(a) != list.get(b))
                return false;
            a++;
            b--;
        }
        return true;

    }
}



//带有输入输出
import dto.ListNode;

import java.util.ArrayList;
import java.util.List;

public class hot_19isPalindrome {
    public boolean isPalindrome(ListNode head){
        List<Integer> list = new ArrayList<>();
        ListNode curr = head;
        while (curr != null){
            list.add(curr.val);
            curr = curr.next;
        }
        int a = 0;
        int b = list.size()-1;
        while (a<b){
            if (list.get(a) != list.get(b))
                return false;
            a++;
            b--;
        }
        return true;
    }

    public static void main(String[] args) {
        ListNode headA = new ListNode(1);
        headA.next = new ListNode(2);
        headA.next.next = new ListNode(2);
        headA.next.next.next = new ListNode(1);
        hot_19isPalindrome hot19isPalindrome = new hot_19isPalindrome();
        boolean result = hot19isPalindrome.isPalindrome(headA);
        System.out.println("输出:" + result);
    }
}