1.题目描述

2.思路
方法1:直接用Python封装好的数据结构OrderedDict(兼具哈希表与双向链表的数据结构)。
方法2:哈希表辅以双向链表。
3.代码(Python3)
方法1:
class LRUCache(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)
方法2:
class DLinkedNode:
def __init__(self, key=0, value=0):
self.key = key
self.value = value
self.prev = None
self.next = None
class LRUCache:
def __init__(self, capacity: int):
self.cache = dict()
# 伪头和伪尾
self.head = DLinkedNode()
self.tail = DLinkedNode()
self.head.next = self.tail
self.tail.prev = self.head
self.capacity = capacity
self.size = 0
def get(self, key: int) -> int:
if key not in self.cache:
return -1
node = self.cache[key]
self.move_to_head(node)
return node.value
def put(self, key: int, value: int) -> None:
print(self.size, self.capacity)
if key not in self.cache:
node = DLinkedNode(key, value)
self.cache[key] = node
self.add_to_head(node)
self.size += 1
if self.size > self.capacity:
removed = self.remove_tail()
self.cache.pop(removed.key)
self.size -= 1
else:
node = self.cache[key]
node.value = value
self.move_to_head(node)
def add_to_head(self, node):
node.prev = self.head
node.next = self.head.next
self.head.next.prev = node
self.head.next = node
def remove_node(self, node):
node.prev.next = node.next
node.next.prev = node.prev
def move_to_head(self, node):
self.remove_node(node)
self.add_to_head(node)
def remove_tail(self):
node = self.tail.prev
self.remove_node(node)
return node
4.执行情况
方法1:

方法2:

5.感想
这两个方法都是官方题解给的,我第一次接触这种LRU的题,没能想出来解决办法。