142. Linked List Cycle II

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?

Assuming head to ring position distance is a, two pointer meet distance ring start position B

2 (A + b + mcycle) = a + B + ncycle;

=>a+b = (n-m) Cycle

A = (n-m) cycle-b;

So the first two pointer meet position and head down together, the position that meet again is the cycle start point.

 PublicListNode detectcycle (ListNode head) {if(Head = =NULL)return NULL; varslow =Head; varFast =Head;  while(fast!=NULL&& Fast.next! =NULL) {Slow=Slow.next; Fast=Fast.next.next; if(slow = = fast) Break; }         if(fast==NULL|| Fast.next = =NULL)return NULL; Slow=Head;  while(Slow! =fast) {Slow=Slow.next; Fast=Fast.next; }         returnslow; }    

Scaling issues

Http://www.cnblogs.com/hiddenfox/p/3408931.html

In the online collection of some of the problems related to the question, the idea of a lot of open, summed up as follows:

1. What is the length of the ring?

2. How do I find the first node in the ring (that is, linked list Cycle II)?

3. How to turn a linked list into a single linked list (release ring)?

4. How can I tell if there is an intersection of two single-linked lists? How do I find the first node that intersects?

142. Linked List Cycle II