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146.lru-cache.python3.py
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146.lru-cache.python3.py
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#
# [146] LRU Cache
#
# https://leetcode.com/problems/lru-cache/description/
#
# algorithms
# Hard (21.19%)
# Total Accepted: 213.9K
# Total Submissions: 981.7K
# Testcase Example: '["LRUCache","put","put","get","put","get","put","get","get","get"]\n[[2],[1,1],[2,2],[1],[3,3],[2],[4,4],[1],[3],[4]]'
#
#
# Design and implement a data structure for Least Recently Used (LRU) 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 reached its capacity, it should invalidate the least recently
# used item before inserting a new item.
#
#
# Follow up:
# Could you do both operations in O(1) time complexity?
#
# Example:
#
# LRUCache cache = new LRUCache( 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.put(4, 4); // evicts key 1
# cache.get(1); // returns -1 (not found)
# cache.get(3); // returns 3
# cache.get(4); // returns 4
#
#
#
from collections import OrderedDict
class LRUCache:
def __init__(self, capacity):
"""
:type capacity: int
"""
self._cap = capacity
self._data = OrderedDict()
def get(self, key):
"""
:type key: int
:rtype: int
"""
res = self._data.get(key, -1)
if res != -1:
self._data.move_to_end(key)
return res
def put(self, key, value):
"""
:type key: int
:type value: int
:rtype: void
"""
if key not in self._data:
self._data[key] = value
else:
self._data[key] = value
self._data.move_to_end(key)
while len(self._data) > self._cap:
self._data.popitem(last=False)
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)