LRU算法实现

What

LRU是Least Recently Used的缩写，即最近最少使用，是一种常用的页面置换算法，

选择最近最久未使用的数据予以淘汰。

Source

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

Design ideas

LRU的算法核心是哈希链表

本质就是HashMap+DoubleLinkedList 时间复杂度是O(1)，哈希表+双向链表的结合体

How to implement

Case1

参考LinkedHashMap

依赖JDK

package com.hhf.study.lru;

import java.util.LinkedHashMap;
import java.util.Map;

public class LRUCacheDemo<K,V> extends LinkedHashMap<K, V> {

    private int capacity;//缓存坑位

    public LRUCacheDemo(int capacity) {
        super(capacity,0.75F,false);
        this.capacity = capacity;
    }

    @Override
    protected boolean removeEldestEntry(Map.Entry<K, V> eldest) {
        return super.size() > capacity;
    }

    public static void main(String[] args) {
        LRUCacheDemo lruCacheDemo = new LRUCacheDemo(3);

        lruCacheDemo.put(1,"a");
        lruCacheDemo.put(2,"b");
        lruCacheDemo.put(3,"c");
        System.out.println(lruCacheDemo.keySet());

        lruCacheDemo.put(4,"d");
        System.out.println(lruCacheDemo.keySet());

        lruCacheDemo.put(3,"c");
        System.out.println(lruCacheDemo.keySet());
        lruCacheDemo.put(3,"c");
        System.out.println(lruCacheDemo.keySet());
        lruCacheDemo.put(3,"c");
        System.out.println(lruCacheDemo.keySet());
        lruCacheDemo.put(5,"x");
        System.out.println(lruCacheDemo.keySet());
    }
}

/**
 * true
 * [1, 2, 3]
 * [2, 3, 4]
 * [2, 4, 3]
 * [2, 4, 3]
 * [2, 4, 3]
 * [4, 3, 5]
 * */

/**
 * false
 [1, 2, 3]
 [2, 3, 4]
 [2, 3, 4]
 [2, 3, 4]
 [2, 3, 4]
 [3, 4, 5]
 */

Case2

不依赖JDK

package com.ming.brush.leetcode.linkedlist;


import org.w3c.dom.Node;

import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;

public class LRUCache2 {


    //map负责查找，构建一个虚拟的双向链表，它里面安装的就是一个个Node节点，作为数据载体。

    //1.构造一个node节点作为数据载体
    class Node<K, V> {
        K key;
        V value;
        Node<K, V> prev;
        Node<K, V> next;

        public Node() {
            this.prev = this.next = null;
        }

        public Node(K key, V value) {
            this.key = key;
            this.value = value;
            this.prev = this.next = null;
        }

    }

    //2 构建一个虚拟的双向链表,,里面安放的就是我们的Node
    class DoubleLinkedList<K, V> {
        Node<K, V> head;
        Node<K, V> tail;

        public DoubleLinkedList() {
            head = new Node<>();
            tail = new Node<>();
            head.next = tail;
            tail.prev = head;
        }

        //3. 添加到头
        public void addHead(Node<K, V> node) {
            node.next = head.next;
            node.prev = head;
            head.next.prev = node;
            head.next = node;
        }

        //4.删除节点
        public void removeNode(Node<K, V> node) {
            node.next.prev = node.prev;
            node.prev.next = node.next;
            node.prev = null;
            node.next = null;
        }

        //5.获得最后一个节点
        public Node getLast() {
            return tail.prev;
        }
    }

    private int cacheSize;
    Map<Integer, Node<Integer, Integer>> map;
    DoubleLinkedList<Integer, Integer> doubleLinkedList;

    public LRUCache2(int cacheSize) {
        this.cacheSize = cacheSize;//坑位
        map = new HashMap<>();//查找
        doubleLinkedList = new DoubleLinkedList<>();
    }

    public int get(int key) {
        if (!map.containsKey(key)) {
            return -1;
        }

        Node<Integer, Integer> node = map.get(key);
        doubleLinkedList.removeNode(node);
        doubleLinkedList.addHead(node);

        return node.value;
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {  //update
            Node<Integer, Integer> node = map.get(key);
            node.value = value;
            map.put(key, node);

            doubleLinkedList.removeNode(node);
            doubleLinkedList.addHead(node);
        } else {
            if (map.size() == cacheSize)  //坑位满了
            {
                Node<Integer, Integer> lastNode = doubleLinkedList.getLast();
                map.remove(lastNode.key);
                doubleLinkedList.removeNode(lastNode);
            }

            //新增一个
            Node<Integer, Integer> newNode = new Node<>(key, value);
            map.put(key, newNode);
            doubleLinkedList.addHead(newNode);

        }
    }

    public static void main(String[] args) {

        LRUCache2 lruCacheDemo = new LRUCache2(3);

        lruCacheDemo.put(1, 1);
        lruCacheDemo.put(2, 2);
        lruCacheDemo.put(3, 3);
        System.out.println(lruCacheDemo.map.keySet());

        lruCacheDemo.put(4, 1);
        System.out.println(lruCacheDemo.map.keySet());

        lruCacheDemo.put(3, 1);
        System.out.println(lruCacheDemo.map.keySet());
        lruCacheDemo.put(3, 1);
        System.out.println(lruCacheDemo.map.keySet());
        lruCacheDemo.put(3, 1);
        System.out.println(lruCacheDemo.map.keySet());
        lruCacheDemo.get(2);
        lruCacheDemo.put(5, 1);
        System.out.println(lruCacheDemo.map.keySet());

    }
}
/*
[1, 2, 3]
[2, 3, 4]
[2, 3, 4]
[2, 3, 4]
[2, 3, 4]
[2, 3, 5]
 */

赏

Thanks

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