嗯嗯,第一眼看见这个题目感觉是用循环顺序表或者队列和哈希表来实现,可是尝试了之后发现一个问题,就是题目中的get操作也会使得某个值被操作从而改变其在顺序表/队列中的位置,这会导致每次get操作时都要按顺序将后半部分的元素往前位移之类的操作,从而时间复杂度为O(N)与题意不符合。所以我就考虑应该用链表加哈希表来实现,不过由于不太熟练就直接看了别人的解答,这题可以使用python里已有的结合了双向链表和哈希表的数据结构如OrderedDict(见实现2),当然这样就和题意不符了。所以这题需要自己将双向链表写出来(见实现1),这个实现的各项操作中,访问哈希表的时间复杂度为O(1),在双向链表的头部添加节点、在双向链表的尾部删除节点的复杂度也为O(1)。而将一个节点移到双向链表的头部,可以分成”删除该节点”和”在双向链表的头部添加节点”两步操作,都可以在O(1) 时间内完成。
class DLNode:
def __init__(self, key=0, value=0):
self.value = value
self.key = key
self.prev = None
self.next = None
class LRUCache1:
def __init__(self, capacity: int):
self.cache = dict()
self.head = DLNode()
self.rear = DLNode()
self.head.next = self.rear
self.rear.prev = self.head
self.cap = capacity
self.num = 0
def get(self, key: int) -> int:
if key not in self.cache:
return -1
else:
node = self.cache[key]
self.moveToHead(node)
return node.value
def put(self, key: int, value: int) -> None:
if key not in self.cache:
node = DLNode(key, value)
self.cache[key] = node
self.addToHead(node)
self.num += 1
if self.num > self.cap:
removed = self.removeRear()
self.cache.pop(removed.key)
self.num -= 1
else:
node = self.cache[key]
node.value = value
self.moveToHead(node)
def removeRear(self):
node = self.rear.prev
self.removeNode(node)
return node
def addToHead(self, node):
node.prev = self.head
node.next = self.head.next
self.head.next.prev = node
self.head.next = node
def moveToHead(self, node):
self.removeNode(node)
self.addToHead(node)
def removeNode(self, node):
node.prev.next = node.next
node.next.prev = node.prev
class LRUCache2(collections.OrderedDict):
def __init__(self, capacity: int):
super().__init__()
self.capacity = capacity
def get(self, key: int) -> int:
if key not in self:
return -1
self.move_to_end(key)
return self[key]
def put(self, key: int, value: int) -> None:
if key in self:
self.move_to_end(key)
self[key] = value
if len(self) > self.capacity:
self.popitem(last=False)
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)
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