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?

思路：由【Leetcode】Linked
List Cycle可知。利用一快一慢两个指针可以推断出链表是否存在环路。

如果两个指针相遇之前slow走了s步，则fast走了2s步。而且fast已经在长度为r的环路中走了n圈，则可知：s = n * r。假定链表长为l。从链表头到环入口点距离为x。从环入口点到相遇点距离为a，则：x + a = s = n * r = (n - 1) * r + r =  (n
- 1) * r + l - x。因此x = (n - 1) * r + (l - x - a)。(l - x - a)为从相遇点到环入口的距离，这意味着当一个指针从链表头出发时，还有一个指针从相遇点開始出发，两者一定会在环入口处相遇。

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

class Solution {

public:

    ListNode *detectCycle(ListNode *head) {

        ListNode * slow = head;

        ListNode * fast = head;

        while(fast && fast->next)

        {

            slow = slow->next;

            fast = fast->next->next;

            if(slow == fast)

            {

                ListNode *slow2 = head;

                while (slow2 != slow)

                {

                    slow2 = slow2->next;

                    slow = slow->next;

                }

                return slow2;

            }

        }

        return NULL;

    }

};