OOD Interview Questions
设计 LRU cache
LRU cache面向对象设计示例
这份 notebook 由 Donne Martin 准备。Source 和 license info 在 GitHub。
设计 LRU cache
Constraints & assumptions
- 我们在 cache 什么?
- cache web queries 的结果
- 可以假设输入都是 valid,还是需要 validate?
- Assume they're valid
- 可以假设能放进 memory 吗?
- Yes
Solution
%%writefile lru_cache.py
class Node(object):
def __init__(self, results):
self.results = results
self.prev = None
self.next = None
class LinkedList(object):
def __init__(self):
self.head = None
self.tail = None
def move_to_front(self, node): # ...
def append_to_front(self, node): # ...
def remove_from_tail(self): # ...
class Cache(object):
def __init__(self, MAX_SIZE):
self.MAX_SIZE = MAX_SIZE
self.size = 0
self.lookup = {} # key: query, value: node
self.linked_list = LinkedList()
def get(self, query)
"""Get the stored query result from the cache.
Accessing a node updates its position to the front of the LRU list.
"""
node = self.lookup.get(query)
if node is None:
return None
self.linked_list.move_to_front(node)
return node.results
def set(self, results, query):
"""Set the result for the given query key in the cache.
When updating an entry, updates its position to the front of the LRU list.
If the entry is new and the cache is at capacity, removes the oldest entry
before the new entry is added.
"""
node = self.lookup.get(query)
if node is not None:
# Key exists in cache, update the value
node.results = results
self.linked_list.move_to_front(node)
else:
# Key does not exist in cache
if self.size == self.MAX_SIZE:
# Remove the oldest entry from the linked list and lookup
self.lookup.pop(self.linked_list.tail.query, None)
self.linked_list.remove_from_tail()
else:
self.size += 1
# Add the new key and value
new_node = Node(results)
self.linked_list.append_to_front(new_node)
self.lookup[query] = new_node