LRU

Java实现最近最少使用

package com.leetcode.design;

import java.util.Hashtable;

public class LRU {
    private Hashtable<Integer,DLinkedNode> cache = new Hashtable<Integer,DLinkedNode>();
    private int count;
    private int capacity;
    private DLinkedNode head,tail;

    public LRU(int capacity){
        this.capacity = capacity;
        this.count = 0;
        head = new DLinkedNode();
        head.pre = null;
        tail = new DLinkedNode();
        tail.post = null;

        head.post = tail;
        tail.pre = head;
    }

    public int get(int key){
        DLinkedNode node = cache.get(key);
        if(node == null){
            return -1;
        }
        this.removeToHead(node);
        return node.value;
    }

    public void set(int key,int value){
        DLinkedNode node = cache.get(key);
        if(node == null){
            DLinkedNode newNode = new DLinkedNode();
            newNode.key = key;
            newNode.value = value;
            this.cache.put(key,newNode);
            this.addNode(newNode);
            ++count;
            if(count>capacity){
                DLinkedNode tail = this.popTail();
                this.cache.remove(tail.key);
                --count;
            }
        }else{
            node.value = value;
            this.removeToHead(node);
        }
    }

    private DLinkedNode popTail() {
        DLinkedNode res = tail.pre;
        this.removeNode(res);
        return res;
    }

    private void removeToHead(DLinkedNode node) {
        this.removeNode(node);
        this.addNode(node);
    }

    private void removeNode(DLinkedNode node) {
        DLinkedNode pre = node.pre;
        DLinkedNode post = node.post;
        pre.post = post;
        post.pre = pre;
    }

    private void addNode(DLinkedNode node) {
        node.pre = head;
        node.post = head.post;

        head.post.pre = node;
        head.post = node;
    }
    
}

创建一个hashtable的缓存，并在构造函数中初始化头节点、尾节点和最大容量，每获取一次存在的节点，均将该节点移动到头部，每加入一个节点，首先判断是否存在该节点，若存在则put替换旧的节点，计数加一，并将其移动到头部，然后判断是否超过最大容量，若超过则移除尾部节点，计数减一