Gidhub BE Developer

# LeetCode : 146. LRU Cache

2021-03-14
goodGid

## Problem

• Need to Retry
``````Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Could you do get and put in O(1) time complexity?
``````

## Example

``````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.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4
``````

## [1] Code (21. 03. 14)

``````class LRUCache {
private final Map<Integer, Integer> map;
private final int capacity;

public LRUCache(int capacity) {
map = new LinkedHashMap<>(capacity, 0.75f, true);
this.capacity = capacity;
}

public int get(int key) {
return map.getOrDefault(key, -1);
}

public void put(int key, int value) {
map.put(key, value);

if (map.size() > capacity) {
int leastUsedKey = map.keySet().iterator().next();
map.remove(leastUsedKey);
}
}
}
``````

Check Point

• LRU Cache를 구현한다.

Algorithm Description

• 직접 구현하는 방법이 있겠지만

Java에서 제공하는 Method를 사용하는 방식이다.

나중엔 직접 구현해서 다시 풀어도 좋을 듯싶다.