题目： Given a linked list, determine if it has a cycle in it.
思路分析： 利用快慢指针slow，fast。 slow指针每次走一步，fast指针每次走两步，倘若存在环，则slow和fast必定在某一时刻相遇。
C++参考代码：
/** * Definition for singly-linked list. * struct ListNode { *int val; *ListNode *next; *ListNode(int x) : val(x), next(NULL) {} * }; */class Solution{public:bool hasCycle(ListNode *head){if (!head) return false;ListNode *slow = head;ListNode *fast = head->next;while (fast && fast->next){if (slow == fast) return true;slow = slow->next;fast = fast->next->next;}return false;}};
C#参考代码：
/** * Definition for singly-linked list. * public class ListNode { *public int val; *public ListNode next; *public ListNode(int x) { *val = x; *next = null; *} * } */{public bool HasCycle(ListNode head){if (head == null) return false;ListNode slow = head;ListNode fast = head.next;while (fast != null && fast.next != null){if (slow == fast) return true;slow = slow.next;fast = fast.next.next;}return false;}}
，也会有想更换的念头。旅行，正好提供了这样一个机会，
