使用 LinkedHashMap 实现 LRU 算法

LRU是Least Recently Used 的缩写，翻译过来就是“最近最少使用”，LRU缓存就是使用这种原理实现，简单的说就是缓存一定量的数据，当超过设定的阈值时就把一些过期的数据删除掉，比如我们缓存10000条数据，当数据小于10000时可以随意添加，当超过10000时就需要把新的数据添加进来，同时要把过期数据删除，以确保我们最大缓存10000条。

LinkedHashMap自身已经实现了LRU功能，请看如下的LinkedHashMap.java源码

    /**     * The iteration ordering method for this linked hash map: <tt>true</tt>     * for access-order, <tt>false</tt> for insertion-order.     *     * @serial     */    final boolean accessOrder;

  /**     * Returns <tt>true</tt> if this map should remove its eldest entry.     * This method is invoked by <tt>put</tt> and <tt>putAll</tt> after     * inserting a new entry into the map.  It provides the implementor     * with the opportunity to remove the eldest entry each time a new one     * is added.  This is useful if the map represents a cache: it allows     * the map to reduce memory consumption by deleting stale entries.     *     * <p>Sample use: this override will allow the map to grow up to 100     * entries and then delete the eldest entry each time a new entry is     * added, maintaining a steady state of 100 entries.     * <pre>     *     private static final int MAX_ENTRIES = 100;     *     *     protected boolean removeEldestEntry(Map.Entry eldest) {     *        return size() &gt; MAX_ENTRIES;     *     }     * </pre>     *     * <p>This method typically does not modify the map in any way,     * instead allowing the map to modify itself as directed by its     * return value.  It <i>is</i> permitted for this method to modify     * the map directly, but if it does so, it <i>must</i> return     * <tt>false</tt> (indicating that the map should not attempt any     * further modification).  The effects of returning <tt>true</tt>     * after modifying the map from within this method are unspecified.     *     * <p>This implementation merely returns <tt>false</tt> (so that this     * map acts like a normal map - the eldest element is never removed).     *     * @param    eldest The least recently inserted entry in the map, or if     *           this is an access-ordered map, the least recently accessed     *           entry.  This is the entry that will be removed it this     *           method returns <tt>true</tt>.  If the map was empty prior     *           to the <tt>put</tt> or <tt>putAll</tt> invocation resulting     *           in this invocation, this will be the entry that was just     *           inserted; in other words, if the map contains a single     *           entry, the eldest entry is also the newest.     * @return   <tt>true</tt> if the eldest entry should be removed     *           from the map; <tt>false</tt> if it should be retained.     */    protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {        return false;    }

现在我们要做的就是两件事：
- 设置accessOrder为true
- 重写removeEldestEntry方法，让它在一定条件下返回true，这样当LinkedHashMap内容填满时，新来的值会挤掉LRU的值。

public class Main {    static public class LRULinkedHashMap<K, V> extends LinkedHashMap<K, V> {        private int capacity;        LRULinkedHashMap(int capacity){            super(16, 0.75f, true);            this.capacity = capacity;        }        @Override        public boolean removeEldestEntry(Map.Entry<K, V> eldest){            return size() > capacity;        }    }    public static void main(String[] args) {        LRULinkedHashMap<String, String> map = new LRULinkedHashMap<>(4);        map.put("rcx1", "a1");  //queue: rcx1        map.put("rcx2", "a2");  //queue: rcx2-rcx1        map.put("rcx3", "a3");  //queue: rcx3-rcx2-rcx1        //map.get("rcx1");  //queue will be changed to: rcx1-rcx3-rcx2        map.put("rcx4", "a4");  //queue: rcx4-rcx3-rcx2-rcx1        map.put("rcx5", "a5");  //queue: rcx5-rcx4-rcx3-rcx2, rcx1 will be removed since the capacity is 4.        System.out.println(map.get("rcx1"));        LinkedHashMap<String, String> map2 = new LinkedHashMap<>(4);        map2.put("rcx1", "a1");        map2.put("rcx2", "a2");        map2.put("rcx3", "a3");        map2.put("rcx4", "a4");        map2.put("rcx5", "a5");        System.out.println(map2.get("rcx1"));    }}

输出结果如下：

null
a1