[LeetCode-Medium] LRU Cache

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.

Follow up:
Could you do get and put in O(1) 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:

• 1 <= capacity <= 3000
• 0 <= key <= 3000
• 0 <= value <= 104
• At most 3 * 104 calls will be made to get and put.

這個題目要我們實踐一個 LRU Cache 的算法

常見的快取演算法有

Solution - Javascript

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) {
    this.cache.delete(key);
    this.cache.set(key, value);
    if (this.cache.size > this.capacity) 
        this.cache.delete(this.cache.keys().next().value);
};