判断链表中是否有环 -- 每日一题04

'''
定义俩个步长不一样的指针，一个每次走一步，一个每次走俩步
如果快慢指针相遇则表示链表中有环，如果没有相遇快指针就走到了链表末尾，那么表示链表没有环
'''
class Solution:
    def hasCycle(self , head: ListNode) -> bool:
        slow = head
        fast = head
        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                break
        if fast is None or fast.next is None:
            return False
        return True