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能够判断出链表是否存在环路。假设两个指针相遇之前slow走了s步，则fast走了2s步，并且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，另一个指针从相遇点开始出发，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;
            
    }
};