java如何实现一个LRU Cache

如何设计实现一个LRU Cache?

目录

  1. 1. 什么是LRU Cache?
  2. 2.实现思路

1. 什么是LRU Cache?

之前,在LeetCode上看到一个LRU Cache实现的题目,题目描述是这样的:

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(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.

简单的说,就是保证基本的get和set的功能的同时,还要保证最近访问(get或put)的节点保持在限定容量的Cache中,如果超过容量则应该把LRU(近期最少使用)的节点删除掉。

那么我们思考一个问题:如何设计实现一个LRU Cache?
那么,我们可能需要使用类似这样的数据结构去实现这个LRU Cache:

这不就是LinkedHashMap吗!
这样做的好处是,getset在不冲突的情况下可以保证O(1)的复杂度,同时,也可以通过双向链表来保证LRU的删除更新操作也能保证O(1)的复杂度。

2.实现思路

在学习了HashMap(#7 )和LinkedHashMap(#8 )后,是不是觉得这俩数据结构简直太适合做LRU Cache了!那么动手实现一下:
基于HashMap和双向链表的实现

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public class LRUCache {
    class Node {
    	Node pre;
    	Node next;
    	Integer key;
    	Integer val;
    	
    	Node(Integer k, Integer v) {
    		key = k;
    		val = v;
    	}
    }
    
    Map map = new HashMap();
    // The head (eldest) of the doubly linked list.
    Node head;
    // The tail (youngest) of the doubly linked list.
    Node tail;
    int cap;
    public LRUCache(int capacity) {
        cap = capacity;
        head = new Node(null, null);
        tail = new Node(null, null);
        head.next = tail;
        tail.pre = head;
    }
    
    public int get(int key) {
        Node n = map.get(key);
        if(n!=null) {
        	n.pre.next = n.next;
        	n.next.pre = n.pre;
        	appendTail(n);
        	return n.val;
        }
        return -1;
    }
    
    public void set(int key, int value) {
        Node n = map.get(key);
        // existed
        if(n!=null) {
	        n.val = value;
	        map.put(key, n);
        	n.pre.next = n.next;
        	n.next.pre = n.pre;
        	appendTail(n);
        	return;
        }
        // else {
        if(map.size() == cap) {
        	Node tmp = head.next;
        	head.next = head.next.next;
        	head.next.pre = head;
        	map.remove(tmp.key);
        }
        n = new Node(key, value);
        // youngest node append taill
        appendTail(n);
        map.put(key, n);
    }

    private void appendTail(Node n) {
    	n.next = tail;
    	n.pre = tail.pre;
    	tail.pre.next = n;
    	tail.pre = n;
    }
}

基于LinkedHashMap的实现
HashMap+双向链表?这不就是LinkedHashMap吗!

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public class LRUCache {
    
    private int capacity;
    private Map cache;
    
    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.cache = new java.util.LinkedHashMap (capacity, 0.75f, true) {
            // 定义put后的移除规则,大于容量就删除eldest
            protected boolean removeEldestEntry(Map.Entry eldest) {
                return size() > capacity;
            }
        };
    }
    
    public int get(int key) {
        if (cache.containsKey(key)) {
            return cache.get(key);
        } else
            return -1;
    }
    
    public void set(int key, int value) {
        cache.put(key, value);
    }
}

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