LeetCode 234. Palindrome Linked List （回文链表）

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?

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题目标签：Linked List

　　题目给了我们一个 linked list，让我们判断它是不是回文。

　　这里可以利用 #206 Reverse Linked List 把右边一半的链表 倒转，然后从左右两头开始比较链表是否是回文。

　　这样的话，首先要找到链表的中间点，然后开始倒转右半边链表。

　　可以利用slow fast pointers 来找到中间点，当fast 走完的时候，slow 正好在中间。

Java Solution:

Runtime beats 39.01%

完成日期：06/11/2017

关键词：singly-linked list

关键点：利用slow fast 指针来找到中间点

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 class Solution
 {
     public boolean isPalindrome(ListNode head)
     {
         ListNode slow = head;
         ListNode fast = head;
         ListNode tail;
 
         // find the middle
         while(fast != null && fast.next != null)
         {
             slow = slow.next;
             fast = fast.next.next;
         }
 
         if(fast != null) // odd length
             slow = slow.next; // move middle to right half
 
         // reverse right half
         tail = reverse(slow);
 
         // compare head and tail
         while(tail != null)
         {
             if(head.val != tail.val)
                 return false;
 
             head = head.next;
             tail = tail.next;
         }
 
         return true;
     }
 
     private ListNode reverse(ListNode cursor)
     {
         ListNode pre = null;
         ListNode next;
 
         while(cursor != null)
         {
             next = cursor.next;
             cursor.next = pre;
             pre = cursor;
             cursor = next;
         }
 
         return pre; // tail node
     }
 }

参考资料：https://discuss.leetcode.com/topic/33376/java-easy-to-understand

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