题解 | #判断链表中是否有环#

判断指针是否有环：

1. 指定两个指针，快指针和慢指针，快指针每次前进两步，慢指针每次前进一步，当快指针与慢指针相遇时有环，当快指针和慢指针不相遇时无环
2. 边界条件，当输入为空时，默认无环

class Solution:
    def hasCycle(self , head: ListNode) -> bool:
        if head is None:
            return False
        fast = head
        slow = head
        while fast is not None and fast.next is not None:
            print(fast.val)
            fast = fast.next.next 
            slow = slow.next 
            if fast == slow:
                return True
        return False