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;
}