Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
思路：采用“快慢指针”查检查链表是否含有环。让一个指针一次走一步，另一个一次走两步，如果链表中含有环，快的指针会再次和慢的指针相遇。
class Solution { public: bool hasCycle(ListNode *head) { ListNode* slow = head; ListNode* fast = head; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) return true; } return false; } };
141. Linked List Cycle (List; Two-Pointers)
原文：http://www.cnblogs.com/qionglouyuyu/p/4853495.html