题目：
题解：
python
class Node:
def __init__(self, key="", count=0):
self.prev = None
self.next = None
self.keys = {key}
self.count = count
def insert(self, node: 'Node') -> 'Node': # 在 self 后插入 node
node.prev = self
node.next = self.next
node.prev.next = node
node.next.prev = node
return node
def remove(self): # 从链表中移除 self
self.prev.next = self.next
self.next.prev = self.prev
class AllOne:
def __init__(self):
self.root = Node()
self.root.prev = self.root
self.root.next = self.root # 初始化链表哨兵，下面判断节点的 next 若为 self.root，则表示 next 为空（prev 同理）
self.nodes = {}
def inc(self, key: str) -> None:
if key not in self.nodes: # key 不在链表中
if self.root.next is self.root or self.root.next.count > 1:
self.nodes[key] = self.root.insert(Node(key, 1))
else:
self.root.next.keys.add(key)
self.nodes[key] = self.root.next
else:
cur = self.nodes[key]
nxt = cur.next
if nxt is self.root or nxt.count > cur.count + 1:
self.nodes[key] = cur.insert(Node(key, cur.count + 1))
else:
nxt.keys.add(key)
self.nodes[key] = nxt
cur.keys.remove(key)
if len(cur.keys) == 0:
cur.remove()
def dec(self, key: str) -> None:
cur = self.nodes[key]
if cur.count == 1: # key 仅出现一次，将其移出 nodes
del self.nodes[key]
else:
pre = cur.prev
if pre is self.root or pre.count < cur.count - 1:
self.nodes[key] = cur.prev.insert(Node(key, cur.count - 1))
else:
pre.keys.add(key)
self.nodes[key] = pre
cur.keys.remove(key)
if len(cur.keys) == 0:
cur.remove()
def getMaxKey(self) -> str:
return next(iter(self.root.prev.keys)) if self.root.prev is not self.root else ""
def getMinKey(self) -> str:
return next(iter(self.root.next.keys)) if self.root.next is not self.root else ""