leetcode--Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:
Can you solve it without using extra space?

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1. /** 
2.  * Definition for singly-linked list. 
3.  * class ListNode { 
4.  *     int val; 
5.  *     ListNode next; 
6.  *     ListNode(int x) { 
7.  *         val = x; 
8.  *         next = null; 
9.  *     } 
10.  * } 
11.  */  
12. public class Solution {  
13.     public ListNode detectCycle(ListNode head) {  
14.         if(null == head || null == head.next) return null;  
15.         ListNode slow = head;  
16.         ListNode fast = head;  
17.         while(slow != null && fast != null && fast.next != null){  
18.             slow = slow.next;  
19.             fast = fast.next.next;  
20.             if(slow == fast){  
21.                 slow = head;  
22.                 while(slow != fast){  
23.                     slow = slow.next;  
24.                     fast = fast.next;  
25.                 }  
26.                 return slow;  
27.             }  
28.         }  
29.         return null;  
30.     }  
31. } 

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46563883