4

#yyds干货盘点# leetcode算法题:回文链表

 2 years ago
source link: https://blog.51cto.com/u_13321676/5576971
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
neoserver,ios ssh client

#yyds干货盘点# leetcode算法题:回文链表

原创

灰太狼_cxh 2022-08-15 11:23:35 博主文章分类:leetcode ©著作权

文章标签 链表 代码实现 文章分类 Java 编程语言 阅读数194

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 。

输入:head = [1,2,2,1]

输出:true

输入:head = [1,2]

输出:false

代码实现:

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isPalindrome(ListNode head) {
if(head == null){
return true;
}
ListNode firstHalfEnd = endOfFirstHalf(head);
ListNode secondHalfStart = reverseList(firstHalfEnd.next);
ListNode p1 = head;
ListNode p2 = secondHalfStart;
boolean result = true;
while(result && p2 != null){
if(p1.val != p2.val){
result = false;
}
p1 = p1.next;
p2 = p2.next;
}
firstHalfEnd.next = reverseList(secondHalfStart);
return result;
}
private ListNode reverseList(ListNode head){
ListNode prev = null;
ListNode curr = head;
while(curr != null){
ListNode nextTemp = curr.next;
curr.next = prev;
prev = curr;
curr = nextTemp;
}
return prev;
}
private ListNode endOfFirstHalf(ListNode head){
ListNode fast = head;
ListNode slow = head;
while(fast.next != null && fast.next.next != null){
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}
  • 收藏
  • 评论
  • 分享
  • 举报

About Joyk


Aggregate valuable and interesting links.
Joyk means Joy of geeK