题目描述

• 写这道题前最好把206.翻转链表 写了
• 有空间复杂度的话都好说，不管是新建链表、还是用字符串equals都好做。
  

思路 & 算法

• 快慢指针：

1. 快指针找终点，慢指针反转前半个链表
2. 快指针回到慢指针的位置（中点，记得进行奇偶判断）
3. 此刻，链表已经被分成两条以slow、fast开始的链表，遍历对比即可

/*** 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) {// 快慢指针，先找到快指针起点ListNode fast = head, slow = head;ListNode temp,pre = null;while(fast != null && fast.next != null){fast = fast.next.next;// 翻转链表temp = slow.next;slow.next = pre;pre = slow;slow = temp;}fast = fast == null ? slow : slow.next;slow = pre;// 奇数 & 偶数// 快慢指针分头行动，一边走一边对while(slow != null){if(slow.val != fast.val){return false;}slow = slow.next;fast = fast.next;}return true;}
}

• 更新

class Solution {public boolean isPalindrome(ListNode head) {if(head.next == null) {return true;}ListNode slow = head, fast = head;ListNode pre = null;// 奇数(fast.next == null)、偶数(fast == null)while(fast != null && fast.next != null) {fast = fast.next.next;ListNode next = slow.next;slow.next = pre;pre = slow;slow = next;}if(fast != null) {slow = slow.next;}while(pre != null) {if(pre.val != slow.val) {return false;}pre = pre.next;slow = slow.next;}return true;}
}