2025/10/16
思路:我只能用标记来判断有没有重新访问过,像 DFS 那样。不标记的办法想不出来
推荐答案:快慢指针(我发现很多链表题都是这么干的),注意:快慢指针相遇的点正好是环入点,可以算出来
2025/10/19
def GetCircleEntrance(head: Node) -> Node:
if head == None:
return False
elif head.next == None:
return False
elif head.next.next == None:
return False
fast = head
slow = head
while fast != None:
fast = fast.next.next
slow = slow.next
if fast == slow:
fast = head
while fast != slow:
fast = fast.next
slow = slow.next
return fast
return False正确