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?

当快指针追上慢指针时，一个指针从head开始，一个指针从追上的位置开始，

每次走一步，相遇是就是环的开始节点。

public ListNode detectCycle(ListNode head) {
	if (head == null) 
		return null;
	ListNode fast, slow;
	fast = head.next;
	slow = head;
	while (fast != slow) {
		if(fast==null || fast.next==null)
			return null;
		fast = fast.next.next;
		slow = slow.next;
	}
	while(head!=fast.next){
		head=head.next;
		fast=fast.next;
	}
	return head;
}