题解 | #判断链表中是否有环#

快慢指针，注意判断是否为空

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(head == nullptr) 
            return false;
        ListNode *fast = head, *slow = head;
        while(fast != nullptr && fast->next != nullptr)
        {
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow)
                return true;
        }
        return false;
    }
};