# LeetCode 146. LRU Cache

## 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.

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.put(4, 4);    // evicts key 1
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.tail = ListNode(-1, -1)

def get(self, key: int) -> int:

node = self.cache.get(key, None)

if not node:
return -1

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.count += 1

if self.count > self.capacity:
tail = self.remove_tail()
del self.cache[tail.key]
self.count -= 1
else:
node.value = value

def remove_node(self, node):
prev = node.prev
new = node.next

prev.next = new
new.prev = prev