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?

 

思想： 从开头到交点的距离是交点到返回来交点的距离的n倍，则从开头出发和头交点出发，则一定会在分叉口碰撞；先找到相交点

1. public ListNode detectCycle(ListNode head) {
2. if(head==null || head.next==null) return null;
3. ListNode slow=head;
4. ListNode fast=head;
5. while(fast!=null && fast.next!=null) {
6. fast=fast.next.next;
7. slow=slow.next;
8. if(fast==slow) break;
9. }
10. if(fast==null || fast.next==null) return null;
11. ListNode p=head;
12. while(p!=fast) {
13. p=p.next;
14. fast=fast.next;
15. }
16. return p;
17. }