LeetCode-LFU_Cache
LeetCode的设计题，比起之前的LRU稍微难点，但是根据《现代操作系统》的解释，需要使用一个counter进行频率的计数，下面给出的并不是O(1)，但是基本思路是对的；原题链接：https://leetcode.com/problems/lfu-cache/
Design and implement a data structure for Least Frequently Used (LFU) cache.
It should support the following operations: get and put.
get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) – Set or insert the value if the key is not already present.
When the cache reaches its capacity, it should invalidate the least frequently used item before inserting a new item.
For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency),
the least recently used key would be evicted.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LFUCache cache = new LFUCache( 2 /* capacity */ );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.get(3); // returns 3.
cache.put(4, 4); // evicts key 1.
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
# _*_ coding: utf-8 _*_
class LFUCache(object):
def __init__(self, capacity):
"""
:type capacity: int
"""
self.capacity = capacity
self.data = {}
self.counter = {}
def get(self, key):
"""
:type key: int
:rtype: int
"""
if key not in self.data:
return -1
count = self.data[key]['used']
self.data[key]['used'] += 1
if count+1 not in self.counter:
self.counter[count+1] = []
self.counter[count].remove(key)
self.counter[count+1].append(key)
return self.data[key]['value']
def put(self, key, value):
"""
:type key: int
:type value: int
:rtype: void
"""
if not self.capacity:
return
if key in self.data:
self.data[key]['value'] = value
count = self.data[key]['used']
self.counter[count].remove(key)
self.counter[count].append(key)
return
if len(self.data) < self.capacity:
self.data[key] = {'value': value, 'used': 0}
if 0 not in self.counter:
self.counter[0] = []
self.counter[0].append(key)
return
i = 0
while True:
if self.counter[i]:
break
else:
i += 1
old_key = self.counter[i].pop(0)
self.data.pop(old_key)
self.data[key] = {'value': value, 'used': 0}
self.counter[0].append(key)
# Your LFUCache object will be instantiated and called as such:
# obj = LFUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)