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。

 class Solution {
     public boolean isPalindrome(ListNode head) {
         ListNode faster= head ,slower= head;
         while(faster!=null && faster.next!=null){
             faster = faster.next.next;
             slower = slower.next;
         }
         if(faster!=null)
             //奇数
             slower=slower.next;
         ListNode newhead = Reverse(slower);
         while(newhead!=null){
             if(head.val!=newhead.val){
                 return false;
             }
             head = head.next;
             newhead = newhead.next;
         }
             return true;

     }

     public ListNode  Reverse(ListNode head) {
         if(head == null ||head.next == null) return head;
         ListNode pre = head;
         head = head.next;
         pre.next = null;
         while(head != null){
             ListNode next = head.next;
             head.next = pre;
             pre = head;
             head = next;
         }
         return pre;
     }

 }