Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

判断一个链表是否存在环，维护快慢指针就可以，如果有环那么快指针一定会追上慢指针，代码如下：

1 class Solution { 2 public: 3     bool hasCycle(ListNode *head) { 4         ListNode * slow, * fast; 5         slow = fast = head; 6         while(slow && fast){ 7             slow = slow->next; 8             fast = fast->next; 9             if(fast) fast = fast->next;10             if(fast && slow && slow == fast)11                 return true;12         }13         return false;14     }15 };