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?

class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *p=findPoint(head),*q=head;
if(!p) return NULL;
while(p!=q){
p=p->next;
q=q->next;
}
return p;

}
//有环的情况，找到第一次相碰点
ListNode *findPoint(ListNode *head){
if(!head) return NULL;
ListNode *fast=head,*slow=head;
while(fast&&fast->next){
slow=slow->next;
fast=fast->next->next;
if(slow==fast) return slow;
}
return NULL;
}
};