LRU

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

JavaScript解法

分析一下题目的意思,这道题是让我们设计一个LRU(最近最少使用)缓存。

  • 调用get方法时查找缓存中对应key的值。

  • 调用put时可以加入或者更新键值。

  • 每次调用getput的时候,都会刷新键值为“最近使用”;而且插入新值的时候如果超出capacity最旧的键值会被舍弃。

  • Follow up 中要求getput的时间复杂度都是O(1)

既然要实现插入和查找都是O(1)复杂度的键值存取,首先直接会想到HashMap。但要同时实现LRU机制,单单一个无序的HashMap是不够的。这里我想到了Java里面的LinkedHashMap,它可以保留hashmap的插入或者访问顺序。那么在JavaScript中怎么实现呢?可以用一个链表来保存键值对的顺序,对于getput已有的值,都是去掉节点再加到结尾;注意如果超出了capacity,就直接去掉代表最旧节点的链表head就行了。为了方便快速在链表中找到已有的值,可以直接在hashmap里面保存链表节点的对象。

实现起来再到修修改改到通过OJ还是费了很多工夫的。

class ListNode {
constructor(key, value) {
this.key = key;
this.val = value;
this.prev = null;
this.next = null;
}
}
class LRUCache {
constructor(capacity) {
this.cap = capacity;
this.map = {};
// 实际保存的键值对数量
this.size = 0;
// 代表最旧的结点
this.head = null;
// 代表最新的结点
this.tail = null;
}
get(key) {
if (!(key in this.map)) {
return -1;
}
let node = this.map[key];
this.put(node.key, node.val);
return node.val;
}
addToTail(node) {
if (this.tail) {
this.tail.next = node;
node.prev = this.tail;
this.tail = node;
} else {
this.tail = this.head = node;
}
}
remove(node) {
if (node.prev) {
node.prev.next = node.next;
} else {
this.head = this.head.next;
}
if (node.next) {
node.next.prev = node.prev;
} else {
this.tail = this.tail.prev;
}
node.prev = node.next = null;
}
put(key, value) {
if (key in this.map) {
let node = this.map[key];
node.val = value;
this.remove(node);
this.addToTail(node);
} else {
let node = new ListNode(key, value);
this.addToTail(node);
this.map[key] = node;
this.size++;
}
// 超出capacity,淘汰最老的
if (this.size > this.cap) {
let key = this.head.key;
this.remove(this.head);
delete this.map[key];
this.size--;
}
}
}

ES6 Map的简单解法

这种解法是之后看到别人的做法:ES6 Javascript, O(1), one Map, fewest lines of code-one-Map-fewest-lines-of-code/155303)

划重点,ES6的Map遍历顺序就是插入顺序!所以这种解法非常简洁,学习一下。

class LRUCache {
constructor(capacity) {
this.capacity = capacity;
this.map = new Map();
}
get(key) {
let val = this.map.get(key);
if (typeof val === 'undefined') { return -1 }
this.map.delete(key);
this.map.set(key, val);
return val;
}
put(key, value) {
if (this.map.has(key)) { this.map.delete(key) }
this.map.set(key, value);
let keys = this.map.keys();
while (this.map.size > this.capacity) { this.map.delete(keys.next().value) }
}
}