题目地址

https://leetcode.com/problems/lru-cache/description/

题目描述

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.

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

思路

由于是保留是最近使用的N条数据，这就和队列的特性很符合， 先进入队列的，先出队列。

因此思路就是用一个队列来记录目前缓存的所有key， 每次push都进行判断，如果 超出最大容量限制则进行清除缓存的操作， 具体清除谁就按照刚才说的队列方式进行处理，同时对key进行入队操作。

get的时候，如果缓存中有，则调整队列（具体操作为删除指定元素和入队两个操作）。 缓存中没有则返回-1

关键点解析

• 队列简化操作

• 队列的操作是这道题的灵魂， 很容易少考虑情况

代码

/*
 * @lc app=leetcode id=146 lang=javascript
 *
 * [146] LRU Cache
 *
 * https://leetcode.com/problems/lru-cache/description/
 *
 * algorithms
 * Hard (24.17%)
 * Total Accepted:    272.8K
 * Total Submissions: 1.1M
 * Testcase Example:  '["LRUCache","put","put","get","put","get","put","get","get","get"]\n[[2],[1,1],[2,2],[1],[3,3],[2],[4,4],[1],[3],[4]]'
 *
 * 
 * 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.
 * 
 * 
 * Follow up:
 * Could you do both operations in O(1) time complexity?
 * 
 * Example:
 * 
 * LRUCache cache = new LRUCache( 2 );
 * 
 * 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
 * 
 * 
 */
/**
 * @param {number} capacity
 */
var LRUCache = function(capacity) {
    this.cache = {};
    this.capacity = capacity;
    this.size = 0;
    this.queue = [];
};

/** 
 * @param {number} key
 * @return {number}
 */
LRUCache.prototype.get = function(key) {
    const hit = this.cache[key];

    if (hit !== undefined) {
        this.queue = this.queue.filter(q => q !== key);
        this.queue.push(key);
        return hit;
    }
    return -1;
};

/** 
 * @param {number} key 
 * @param {number} value
 * @return {void}
 */
LRUCache.prototype.put = function(key, value) {
    const hit = this.cache[key];

    // update cache
    this.cache[key] = value;

    if (!hit) {
        // invalid cache and resize size;
        if (this.size === this.capacity) {
            // invalid cache
            const key = this.queue.shift();
            this.cache[key] = undefined;
        } else {
            this.size = this.size + 1;
        }
        this.queue.push(key);
    } else {
        this.queue = this.queue.filter(q => q !== key);
        this.queue.push(key);
    }
};

/** 
 * 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)
 */