环形链表

解决方式

def hasCycle(self, head: Optional[ListNode]) -> bool:
  fast = slow = head
  while fast and fast.next:
    fast = fast.next.next
    slow = slow.next
    if slow == fast:
      return true
  return false

复杂度

时间复杂度：O(n)
空间复杂度：O(1)