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。
java
//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);
}
}