Leetcode – 146. LRU Cache
Problem
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache class:
LRUCache(int capacity)Initialize the LRU cache with positive sizecapacity.int get(int key)Return the value of thekeyif the key exists, otherwise return-1.void put(int key, int value)Update the value of thekeyif thekeyexists. Otherwise, add thekey-valuepair to the cache. If the number of keys exceeds thecapacityfrom this operation, evict the least recently used key.
The functions get and put must each run in O(1) average time complexity.
Example 1:
Input ["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] Output [null, null, null, 1, null, -1, null, -1, 3, 4] Explanation LRUCache lRUCache = new LRUCache(2); lRUCache.put(1, 1); // cache is {1=1} lRUCache.put(2, 2); // cache is {1=1, 2=2} lRUCache.get(1); // return 1 lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3} lRUCache.get(2); // returns -1 (not found) lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} lRUCache.get(1); // return -1 (not found) lRUCache.get(3); // return 3 lRUCache.get(4); // return 4
Solution
Python has OrderedDict which is suitable for this task
from collections import OrderedDict
class LRUCache:
def __init__(self, capacity: int):
self._capacity = capacity
self._cache = OrderedDict()
def get(self, key: int) -> int:
# key does not exist in cache
if key not in self._cache:
return -1
# move key to the end
self._cache.move_to_end(key)
return self._cache[key]
def put(self, key: int, value: int) -> None:
# move item to end if it already is in the cache
if key in self._cache:
self._cache.move_to_end(key)
self._cache[key] = value
# remove first item if the cache exceeds the capacity
if len(self._cache) > self._capacity:
self._cache.popitem(last=False)Result
