环形链表 II

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