Description
https://leetcode.com/problems/lru-cache/
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.
The cache is initialized with a positive capacity.
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
Explanation
use linked list and hashmap together
Python Solution
class ListNode:
def __init__(self, key, value):
self.key = key
self.value = value
self.next = None
self.prev = None
class LRUCache:
def __init__(self, capacity: int):
self.cache = {}
self.count = 0
self.capacity = capacity
self.head = ListNode(-1, -1)
self.tail = ListNode(-1, -1)
self.head.next = self.tail
self.tail.prev = self.head
def get(self, key: int) -> int:
node = self.cache.get(key, None)
if not node:
return -1
self.insert_to_head(node)
return node.value
def put(self, key: int, value: int) -> None:
node = self.cache.get(key)
if not node:
new_node = ListNode(key, value)
self.cache[key] = new_node
self.add_node(new_node)
self.count += 1
if self.count > self.capacity:
tail = self.remove_tail()
del self.cache[tail.key]
self.count -= 1
else:
node.value = value
self.insert_to_head(node)
def add_node(self, node):
node.prev = self.head
node.next = self.head.next
self.head.next.prev = node
self.head.next = node
def remove_node(self, node):
prev = node.prev
new = node.next
prev.next = new
new.prev = prev
def insert_to_head(self, node):
self.remove_node(node)
self.add_node(node)
def remove_tail(self):
res = self.tail.prev
self.remove_node(res)
return res
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)
- Time complexity: O(N).
- Space complexity: O(N).