数据缓存历险记（四）--LRU 大师兄的 Java 实现

public class LinkedHashMap<K,V>  extends HashMap<K,V> implements Map<K,V>

import java.util.LinkedHashMap;import java.util.Map;
/** * @author lucas * @create 2021-08-09 19:53 * @description LRU算法的基础实现 */public class LRUCacheDemo<k, v> extends LinkedHashMap<k, v> {
    /**     * 1.确定初始容量     * 2.初始化构造器的，容量capacity     * 3.删除操作     */    private int capacity;  //最大容量

    public LRUCacheDemo(int capacity) {        /**         *  @param  initialCapacity the initial capacity         * @param  loadFactor      the load factor         * @param  accessOrder     the ordering mode - <tt>true</tt> for         *         access-order, <tt>false</tt> for insertion-order 访问顺序         */        super(capacity, 0.75F, false);        this.capacity = capacity;    }
    //    默认的构造方法基于，负载因子和顺序//    public LRUCacheDemo(int initialCapacity, float loadFactor, boolean accessOrder, int capacity) {//        super(initialCapacity, loadFactor, accessOrder);//        this.capacity = capacity;//    }
    //删除操作    @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("增加第四个后，列表是"+lruCacheDemo.keySet());
        /**         * [1, 2, 3]         * 增加第四个后，列表是[2, 3, 4]         * 解析： 当前第四个进来之后，4这个值将最近最少使用的直接踢出去         */
        System.out.println("--------------------------------------------");        lruCacheDemo.put(3, "d");        System.out.println(lruCacheDemo.keySet());        lruCacheDemo.put(3, "d");        System.out.println(lruCacheDemo.keySet());        lruCacheDemo.put(3, "d");        System.out.println(lruCacheDemo.keySet());
        /***         * [2, 4, 3]         * [2, 4, 3]         * [2, 4, 3]         * 最近一直使用的，3的值最频繁，放在最右侧         */        System.out.println("--------------------------------------------");        lruCacheDemo.put(5, "d");        System.out.println(lruCacheDemo.keySet());
        /**         * [4, 3, 5]         * 5,进入之后，直接将最少使用的2挤出去         */
        /**  访问顺序（accessOrder）为ture         * [1, 2, 3]         * 增加第四个后，列表是[2, 3, 4]         * --------------------------------------------         * [2, 4, 3]         * [2, 4, 3]         * [2, 4, 3]         * --------------------------------------------         * [4, 3, 5]         */```js

    /**访问顺序（accessOrder）为false     * [1, 2, 3]     * 增加第四个后，列表是[2, 3, 4]     * --------------------------------------------     * [2, 3, 4]     * [2, 3, 4]     * [2, 3, 4]     * --------------------------------------------     * [3, 4, 5]     */}

* 访问顺序accessOreder* - true for access-order,* -false for insertion-order