题解 | #判断链表中是否有环#

用快慢指针解题，如果有环，fast一定会先进入环，而slow后进入环，二者一定能在换上相遇。

 * 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) {
        ListNode *fast = head;
        ListNode *slow = head;
        while(slow!=NULL && fast->next !=NULL && fast->next->next!=NULL)
        {
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow)
            {
                return true;
            }
            
        }
        return false;

    }
};