Implement with hash and double linked list
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song 2019-06-05 09:12:17 +08:00 committed by GitHub
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# Definition for singly-linked list.
class DbListNode(object):
def __init__(self, x, y):
self.key = x
self.val = y
self.next = None
self.prev = None
class LRUCache:
'''
leet code: 146
运用你所掌握的数据结构设计和实现一个 LRU (最近最少使用) 缓存机制
它应该支持以下操作 获取数据 get 写入数据 put
获取数据 get(key) - 如果密钥 (key) 存在于缓存中则获取密钥的值总是正数否则返回 -1
写入数据 put(key, value) - 如果密钥不存在则写入其数据值
当缓存容量达到上限时它应该在写入新数据之前删除最近最少使用的数据值从而为新的数据值留出空间
哈希表+双向链表
哈希表: 查询 O(1)
双向链表: 有序, 增删操作 O(1)
Author: Ben
'''
def __init__(self, capacity: int):
self.cap = capacity
self.hkeys = {}
# self.top和self.tail作为哨兵节点, 避免越界
self.top = DbListNode(None, -1)
self.tail = DbListNode(None, -1)
self.top.next = self.tail
self.tail.prev = self.top
def get(self, key: int) -> int:
if key in self.hkeys.keys():
# 更新结点顺序
cur = self.hkeys[key]
# 跳出原位置
cur.next.prev = cur.prev
cur.prev.next = cur.next
# 最近用过的置于链表首部
top_node = self.top.next
self.top.next = cur
cur.prev = self.top
cur.next = top_node
top_node.prev = cur
return self.hkeys[key].val
return -1
def put(self, key: int, value: int) -> None:
if key in self.hkeys.keys():
cur = self.hkeys[key]
cur.val = value
# 跳出原位置
cur.prev.next = cur.next
cur.next.prev = cur.prev
# 最近用过的置于链表首部
top_node = self.top.next
self.top.next = cur
cur.prev = self.top
cur.next = top_node
top_node.prev = cur
else:
# 增加新结点至首部
cur = DbListNode(key, value)
self.hkeys[key] = cur
# 最近用过的置于链表首部
top_node = self.top.next
self.top.next = cur
cur.prev = self.top
cur.next = top_node
top_node.prev = cur
if len(self.hkeys.keys()) > self.cap:
self.hkeys.pop(self.tail.prev.key)
# 去掉原尾结点
self.tail.prev.prev.next = self.tail
self.tail.prev = self.tail.prev.prev
def __repr__(self):
vals = []
p = self.top.next
while p.next:
vals.append(str(p.val))
p = p.next
return '->'.join(vals)
if __name__ == '__main__':
cache = LRUCache(2)
cache.put(1, 1)
cache.put(2, 2)
print(cache)
cache.get(1) # 返回 1
cache.put(3, 3) # 该操作会使得密钥 2 作废
print(cache)
cache.get(2) # 返回 -1 (未找到)
cache.put(4, 4) # 该操作会使得密钥 1 作废
print(cache)
cache.get(1) # 返回 -1 (未找到)
cache.get(3) # 返回 3
print(cache)
cache.get(4) # 返回 4
print(cache)