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环形链表 II ​

解决方式 ​

python
def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
  fast = slow = head
  while head and head.next:
    slow = slow.next
    fast = fast.next.next
    # 有环则相遇
    if fast == slow:
      slow = head
      while slow != fast:
        slow = slow.next
        fast = fast.next
      return slow
  return None

复杂度 ​

时间复杂度: O(N) 空间复杂度: O(1)