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java实现极简的LRU算法

import java.util.LinkedHashMap;
import java.util.Map;
 
/**
 * LRU (Least Recently Used) 
 */
public class LRUCache<K, V> extends LinkedHashMap<K, V> {
    /**

*/
private static final long serialVersionUID = 1L;
//缓存大小
    private int cacheSize;
 
    public LRUCache(int cacheSize) {
        //第三个参数true是关键
        super(10, 0.75f, true);
        this.cacheSize = cacheSize;
    }
 
    /**
     * 缓存是否已满
     */
    @Override
    protected boolean removeEldestEntry(Map.Entry<K, V> eldest) {
        boolean r = size() > cacheSize;
        if(r){
            System.out.println("清除缓存key:"+eldest.getKey());
        }
        return r;
    }

 

    //测试

    public static void main(String[] args) {
        LRUCache<String, String> cache = new LRUCache<String, String>(5);
        cache.put("1", "1");
        cache.put("2", "2");
        cache.put("3", "3");
        cache.put("4", "4");
        cache.put("5", "5");
 
        System.out.println("初始:");
        System.out.println(cache.keySet());
        System.out.println("访问3:");
        cache.get("3");
        System.out.println(cache.keySet());
        System.out.println("访问2:");
        cache.get("2");
        System.out.println(cache.keySet());
        System.out.println("增加数据6,7:");
        cache.put("6", "6");
        cache.put("7", "7");
        System.out.println(cache.keySet());
    }

}


运行结果如下:

初始化:
[1, 2, 3, 4, 5]
访问3:
[1, 2, 4, 5, 3]
访问2:
[1, 4, 5, 3, 2]
增加数据6,7:
清除缓存key:1
清除缓存key:4
[5, 3, 2, 6, 7]

posted @ 2015-11-02 23:12  程序优化  阅读(1745)  评论(0编辑  收藏  举报