234. Palindrome Linked List(判断链表是否回文)

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

 

思路：

1、 利用快慢指针找到链表中点

2、从中点开始反转链表，判断是否相等。

 

 1 class Solution {
 2     public boolean isPalindrome(ListNode head) {
 3         ListNode faster= head ,slower= head;
 4         while(faster!=null && faster.next!=null){
 5             faster = faster.next.next;
 6             slower = slower.next;
 7         }
 8         if(faster!=null)
 9             //奇数
10             slower=slower.next;
11         ListNode newhead = Reverse(slower);
12         while(newhead!=null){
13             if(head.val!=newhead.val){
14                 return false;
15             }
16             head = head.next;
17             newhead = newhead.next;
18         }
19             return true;
20 
21     }
22 
23     public ListNode  Reverse(ListNode head) {
24         if(head == null ||head.next == null) return head;
25         ListNode pre = head;
26         head = head.next;
27         pre.next = null;
28         while(head != null){
29             ListNode next = head.next;
30             head.next = pre;
31             pre = head;
32             head = next;
33         }
34         return pre;
35     }
36 
37 }