Linked List Cycle II

问题描述：

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

null

.

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?
代码实现：
/**
* 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(head==null) return null;
ListNode fast=head;
ListNode slow=head;
do{
if(fast!=null) fast=fast.next;
if(fast!=null){
fast=fast.next;
}else{
return null;
}
slow=slow.next;
}while(fast != slow);

slow=head;
while(fast!=slow){
fast=fast.next;
slow=slow.next;
}

return slow;
}
}