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() > 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