方法1：哈希表，空间复杂度O（n）

方法2：快慢指针，龟兔赛跑

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        #时间复杂度O（1），龟兔赛跑
        if not head or not head.next: return False
        fast = head.next
        slow = head
        while slow != fast:
            if not fast or not fast.next: return False
            slow = slow.next
            fast = fast.next.next
        return True
        '''
        #时间复杂度O（n）
        seen = set()
        while head:
            if head in seen: return True
            else: seen.add(head)
            head = head.next
        return False
        '''