[LeetCode]Link 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?

思考：第一步：设环长为len，快慢指针q、p相遇时，q比p多走了k*len。
        第二部：p先走k*len步，p，q一起走每次一步，相遇时p比q多走了一个环距离，此时q就是环开始结点。

通过分析AC的感觉真好！

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(!head||!head->next) return NULL;
		ListNode *p=head;int countP=0;
		ListNode *q=p->next;int countQ=1;
		bool flag=false;
		int len=0;
		while(q&&q->next)
		{
			if(p==q)
			{
				len=countQ-countP;
				flag=true;
				break;
			}
			else
			{
				q=q->next->next;
				p=p->next;
				countP+=1;
				countQ+=2;
			}

		}
		if(flag)
		{
			p=head;
			q=head;
			while(len--)
			{
				p=p->next;
			}
			while(p!=q)
			{
				p=p->next;
				q=q->next;
			}
			return p;
		}
		else return NULL;
    }
};