[leetcode] LRU Cache @ Python
[leetcode] LRU Cache @ Python
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
思路分析：可以考虑利用dict数据结构，查找时间是O(1)。但是Python的dictionary缺点是无序。而collections.OrderedDict是有序的, 后加入的元素一定排在先加入元素的后面, 操作和dictionary类似。所以本题目将利用此有序dictionary数据结构来实现。
作为背景知识，请复习：
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import collections
a = collections.OrderedDict()
a[1] = 10 # 若1在a中就更新其值, 若1不在a中就加入(1, 10)这一对key-value。
a[2] = 20
a[3] = 30
del a[2]
a.popitem(last = True) # 弹出尾部的元素
a.popitem(last = False) # 弹出头部的元素
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代码如下：
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class LRUCache:
# @param capacity, an integer
def __init__(self, capacity):
LRUCache.capacity = capacity
LRUCache.length = 0
LRUCache.dict = collections.OrderedDict()
# @return an integer
def get(self, key):
try:
value = LRUCache.dict[key]
del LRUCache.dict[key]
LRUCache.dict[key] = value
return value
except:
return -1
# @param key, an integer
# @param value, an integer
# @return nothing
def set(self, key, value):
try:
del LRUCache.dict[key]
LRUCache.dict[key] = value
except:
if LRUCache.length == LRUCache.capacity:
LRUCache.dict.popitem(last = False)
LRUCache.length -= 1
LRUCache.dict[key] = value
LRUCache.length +=1
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