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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plain copy

/** 

 * Definition for singly-linked list. 

 * class ListNode { 

 *     int val; 

 *     ListNode next; 

 *     ListNode(int x) { 

 *         val = x; 

 *         next = null; 

 *     } 

 * } 

 */  

public class Solution {  

    public ListNode detectCycle(ListNode head) {  

        if(null == head || null == head.next) return null;  

        ListNode slow = head;  

        ListNode fast = head;  

        while(slow != null && fast != null && fast.next != null){  

            slow = slow.next;  

            fast = fast.next.next;  

            if(slow == fast){  

                slow = head;  

                while(slow != fast){  

                    slow = slow.next;  

                    fast = fast.next;  

                }  

                return slow;  

            }  

        }  

        return null;  

    }  

} 

原文链接http://blog.csdn.net/crazy__chen/article/details/46563883