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 size capacity. int get(int key) Return the value of the key if the key exists, otherwise return -1. void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key. The functions get and put must each run in O(1) average time complexity.
class Node:
def __init__(self, val, key):
self.val = val
self.key = key
self.prev = None
self.nxt = None
class DBList:
def __init__(self):
self.dummy = Node(None, None)
self.dummy.nxt = self.dummy
self.dummy.prev = self.dummy
self.size = 0
def remove(self, node):
node.prev.nxt = node.nxt
node.nxt.prev = node.prev
self.size -= 1
def append(self, node):
node.prev = self.dummy.prev
node.nxt = self.dummy
self.dummy.prev.nxt = node
self.dummy.prev = node
self.size += 1
class LRUCache:
def __init__(self, capacity: int):
self.data = {}
self.dblist = DBList()
self.cap = capacity
def get(self, key: int) -> int:
if key not in self.data:
return -1
val = self.data[key].val
self.dblist.remove(self.data[key])
self.dblist.append(self.data[key])
return val
def put(self, key: int, value: int) -> None:
if key in self.data:
self.dblist.remove(self.data[key])
self.data[key].val = value
self.dblist.append(self.data[key])
return
if self.dblist.size == self.cap:
del self.data[self.dblist.dummy.nxt.key]
self.dblist.remove(self.dblist.dummy.nxt)
self.data[key] = Node(value, key)
self.dblist.append(self.data[key])
Related question: 460. LFU Cache