LeetCode-in-All

146. LRU Cache

Medium

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

The functions get and put must each run in O(1) average time complexity.

Example 1:

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.get(2); // returns -1 (not found)

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

lRUCache.get(1); // return -1 (not found)

lRUCache.get(3); // return 3

lRUCache.get(4); // return 4

Constraints:

Solution

/**
 * @param {number} capacity
 */
var LRUCache = function (capacity) {
    this.capacity = capacity
    this.cache = new Map()
};

/** 
 * @param {number} key
 * @return {number}
 */
LRUCache.prototype.get = function (key) {
    if (!this.cache.has(key)) {
        return -1
    }

    const value = this.cache.get(key)
    this.cache.delete(key)
    this.cache.set(key, value)
    return value
};

/** 
 * @param {number} key 
 * @param {number} value
 * @return {void}
 */
LRUCache.prototype.put = function (key, value) {
    if (this.cache.has(key)) {
        this.cache.delete(key)
    } else if (this.cache.size >= this.capacity) {
        const lrukey = this.cache.keys().next().value
        this.cache.delete(lrukey)
    }

    this.cache.set(key, value)
};

/** 
 * Your LRUCache object will be instantiated and called as such:
 * var obj = new LRUCache(capacity)
 * var param_1 = obj.get(key)
 * obj.put(key,value)
 */

export { LRUCache }