[转]LRU缓存介绍与实现 (Java) .

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引子：

我们平时总会有一个电话本记录所有朋友的电话，但是，如果有朋友经常联系，那些朋友的电话号码不用翻电话本我们也能记住，但是，如果长时间没有联系了，要再次联系那位朋友的时候，我们又不得不求助电话本，但是，通过电话本查找还是很费时间的。但是，我们大脑能够记住的东西是一定的，我们只能记住自己最熟悉的，而长时间不熟悉的自然就忘记了。

其实，计算机也用到了同样的一个概念，我们用缓存来存放以前读取的数据，而不是直接丢掉，这样，再次读取的时候，可以直接在缓存里面取，而不用再重新查找一遍，这样系统的反应能力会有很大提高。但是，当我们读取的个数特别大的时候，我们不可能把所有已经读取的数据都放在缓存里，毕竟内存大小是一定的，我们一般把最近常读取的放在缓存里（相当于我们把最近联系的朋友的姓名和电话放在大脑里一样）。现在，我们就来研究这样一种缓存机制。

LRU缓存：

LRU缓存利用了这样的一种思想。LRU是Least Recently Used 的缩写，翻译过来就是“最近最少使用”，也就是说，LRU缓存把最近最少使用的数据移除，让给最新读取的数据。而往往最常读取的，也是读取次数最多的，所以，利用LRU缓存，我们能够提高系统的performance.

实现：

要实现LRU缓存，我们首先要用到一个类 LinkedHashMap。用这个类有两大好处：一是它本身已经实现了按照访问顺序的存储，也就是说，最近读取的会放在最前面，最最不常读取的会放在最后（当然，它也可以实现按照插入顺序存储）。第二，LinkedHashMap本身有一个方法用于判断是否需要移除最不常读取的数，但是，原始方法默认不需要移除（这是，LinkedHashMap相当于一个linkedlist），所以，我们需要override这样一个方法，使得当缓存里存放的数据个数超过规定个数后，就把最不常用的移除掉。LinkedHashMap的API写得很清楚，推荐大家可以先读一下。

要基于LinkedHashMap来实现LRU缓存，我们可以选择inheritance, 也可以选择 delegation，我更喜欢delegation。基于delegation的实现已经有人写出来了，而且写得很漂亮，我就不班门弄斧了。代码如下：

[java] view plain copy print ?

import java.util.LinkedHashMap;
import java.util.Collection;
import java.util.Map;
import java.util.ArrayList;
/**
* An LRU cache, based on <code>LinkedHashMap</code>.
*
*
* This cache has a fixed maximum number of elements (<code>cacheSize</code>).
* If the cache is full and another entry is added, the LRU (least recently used) entry is dropped.
*
*
* This class is thread-safe. All methods of this class are synchronized.
*
*
* Author: Christian d'Heureuse, Inventec Informatik AG, Zurich, Switzerland
* Multi-licensed: EPL / LGPL / GPL / AL / BSD.
*/
public class LRUCache<K,V> {
private static final float hashTableLoadFactor = 0.75f;
private LinkedHashMap<K,V> map;
private int cacheSize;
/**
* Creates a new LRU cache.
* @param cacheSize the maximum number of entries that will be kept in this cache.
*/
public LRUCache (int cacheSize) {
this.cacheSize = cacheSize;
int hashTableCapacity = (int)Math.ceil(cacheSize / hashTableLoadFactor) + 1;
map = new LinkedHashMap<K,V>(hashTableCapacity, hashTableLoadFactor, true) {
// (an anonymous inner class)
private static final long serialVersionUID = 1;
@Override protected boolean removeEldestEntry (Map.Entry<K,V> eldest) {
return size() > LRUCache.this.cacheSize; }}; }
/**
* Retrieves an entry from the cache.
* The retrieved entry becomes the MRU (most recently used) entry.
* @param key the key whose associated value is to be returned.
* @return the value associated to this key, or null if no value with this key exists in the cache.
*/
public synchronized V get (K key) {
return map.get(key); }
/**
* Adds an entry to this cache.
* The new entry becomes the MRU (most recently used) entry.
* If an entry with the specified key already exists in the cache, it is replaced by the new entry.
* If the cache is full, the LRU (least recently used) entry is removed from the cache.
* @param key the key with which the specified value is to be associated.
* @param value a value to be associated with the specified key.
*/
public synchronized void put (K key, V value) {
map.put (key, value); }
/**
* Clears the cache.
*/
public synchronized void clear() {
map.clear(); }
/**
* Returns the number of used entries in the cache.
* @return the number of entries currently in the cache.
*/
public synchronized int usedEntries() {
return map.size(); }
/**
* Returns a <code>Collection</code> that contains a copy of all cache entries.
* @return a <code>Collection</code> with a copy of the cache content.
*/
public synchronized Collection<Map.Entry<K,V>> getAll() {
return new ArrayList<Map.Entry<K,V>>(map.entrySet()); }
} // end class LRUCache
------------------------------------------------------------------------------------------
// Test routine for the LRUCache class.
public static void main (String[] args) {
LRUCache<String,String> c = new LRUCache<String, String>(3);
c.put ("1", "one"); // 1
c.put ("2", "two"); // 2 1
c.put ("3", "three"); // 3 2 1
c.put ("4", "four"); // 4 3 2
if (c.get("2") == null) throw new Error(); // 2 4 3
c.put ("5", "five"); // 5 2 4
c.put ("4", "second four"); // 4 5 2
// Verify cache content.
if (c.usedEntries() != 3) throw new Error();
if (!c.get("4").equals("second four")) throw new Error();
if (!c.get("5").equals("five")) throw new Error();
if (!c.get("2").equals("two")) throw new Error();
// List cache content.
for (Map.Entry<String, String> e : c.getAll())
System.out.println (e.getKey() + " : " + e.getValue()); }

import java.util.LinkedHashMap;
import java.util.Collection;
import java.util.Map;
import java.util.ArrayList;

/**
* An LRU cache, based on <code>LinkedHashMap</code>.
*
* <p>
* This cache has a fixed maximum number of elements (<code>cacheSize</code>).
* If the cache is full and another entry is added, the LRU (least recently used) entry is dropped.
*
* <p>
* This class is thread-safe. All methods of this class are synchronized.
*
* <p>
* Author: Christian d'Heureuse, Inventec Informatik AG, Zurich, Switzerland<br>
* Multi-licensed: EPL / LGPL / GPL / AL / BSD.
*/
public class LRUCache<K,V> {

private static final float   hashTableLoadFactor = 0.75f;

private LinkedHashMap<K,V>   map;
private int                  cacheSize;

/**
* Creates a new LRU cache.
* @param cacheSize the maximum number of entries that will be kept in this cache.
*/
public LRUCache (int cacheSize) {
   this.cacheSize = cacheSize;
   int hashTableCapacity = (int)Math.ceil(cacheSize / hashTableLoadFactor) + 1;
   map = new LinkedHashMap<K,V>(hashTableCapacity, hashTableLoadFactor, true) {
      // (an anonymous inner class)
      private static final long serialVersionUID = 1;
      @Override protected boolean removeEldestEntry (Map.Entry<K,V> eldest) {
         return size() > LRUCache.this.cacheSize; }}; }

/**
* Retrieves an entry from the cache.<br>
* The retrieved entry becomes the MRU (most recently used) entry.
* @param key the key whose associated value is to be returned.
* @return    the value associated to this key, or null if no value with this key exists in the cache.
*/
public synchronized V get (K key) {
   return map.get(key); }

/**
* Adds an entry to this cache.
* The new entry becomes the MRU (most recently used) entry.
* If an entry with the specified key already exists in the cache, it is replaced by the new entry.
* If the cache is full, the LRU (least recently used) entry is removed from the cache.
* @param key    the key with which the specified value is to be associated.
* @param value  a value to be associated with the specified key.
*/
public synchronized void put (K key, V value) {
   map.put (key, value); }

/**
* Clears the cache.
*/
public synchronized void clear() {
   map.clear(); }

/**
* Returns the number of used entries in the cache.
* @return the number of entries currently in the cache.
*/
public synchronized int usedEntries() {
   return map.size(); }

/**
* Returns a <code>Collection</code> that contains a copy of all cache entries.
* @return a <code>Collection</code> with a copy of the cache content.
*/
public synchronized Collection<Map.Entry<K,V>> getAll() {
   return new ArrayList<Map.Entry<K,V>>(map.entrySet()); }

} // end class LRUCache
------------------------------------------------------------------------------------------
// Test routine for the LRUCache class.
public static void main (String[] args) {
   LRUCache<String,String> c = new LRUCache<String, String>(3);
   c.put ("1", "one");                           // 1
   c.put ("2", "two");                           // 2 1
   c.put ("3", "three");                         // 3 2 1
   c.put ("4", "four");                          // 4 3 2
   if (c.get("2") == null) throw new Error();    // 2 4 3
   c.put ("5", "five");                          // 5 2 4
   c.put ("4", "second four");                   // 4 5 2
   // Verify cache content.
   if (c.usedEntries() != 3)              throw new Error();
   if (!c.get("4").equals("second four")) throw new Error();
   if (!c.get("5").equals("five"))        throw new Error();
   if (!c.get("2").equals("two"))         throw new Error();
   // List cache content.
   for (Map.Entry<String, String> e : c.getAll())
      System.out.println (e.getKey() + " : " + e.getValue()); }

代码出自：http://www.source-code.biz/snippets/java/6.htm

在博客 http://gogole.iteye.com/blog/692103 里，作者使用的是双链表 + hashtable 的方式实现的。如果在面试题里考到如何实现LRU，考官一般会要求使用双链表 + hashtable 的方式。所以，我把原文的部分内容摘抄如下：

双链表 + hashtable实现原理：

将Cache的所有位置都用双连表连接起来，当一个位置被命中之后，就将通过调整链表的指向，将该位置调整到链表头的位置，新加入的Cache直接加到链表头中。这样，在多次进行Cache操作后，最近被命中的，就会被向链表头方向移动，而没有命中的，而想链表后面移动，链表尾则表示最近最少使用的Cache。当需要替换内容时候，链表的最后位置就是最少被命中的位置，我们只需要淘汰链表最后的部分即可。

[java] view plain copy print ?

public class LRUCache {
private int cacheSize;
private Hashtable<Object, Entry> nodes;//缓存容器
private int currentSize;
private Entry first;//链表头
private Entry last;//链表尾
public LRUCache(int i) {
currentSize = 0;
cacheSize = i;
nodes = new Hashtable<Object, Entry>(i);//缓存容器
}
/**
* 获取缓存中对象,并把它放在最前面
*/
public Entry get(Object key) {
Entry node = nodes.get(key);
if (node != null) {
moveToHead(node);
return node;
} else {
return null;
}
}
/**
* 添加 entry到hashtable, 并把entry
*/
public void put(Object key, Object value) {
//先查看hashtable是否存在该entry, 如果存在，则只更新其value
Entry node = nodes.get(key);
if (node == null) {
//缓存容器是否已经超过大小.
if (currentSize >= cacheSize) {
nodes.remove(last.key);
removeLast();
} else {
currentSize++;
}
node = new Entry();
}
node.value = value;
//将最新使用的节点放到链表头，表示最新使用的.
moveToHead(node);
nodes.put(key, node);
}
/**
* 将entry删除, 注意：删除操作只有在cache满了才会被执行
*/
public void remove(Object key) {
Entry node = nodes.get(key);
//在链表中删除
if (node != null) {
if (node.prev != null) {
node.prev.next = node.next;
}
if (node.next != null) {
node.next.prev = node.prev;
}
if (last == node)
last = node.prev;
if (first == node)
first = node.next;
}
//在hashtable中删除
nodes.remove(key);
}
/**
* 删除链表尾部节点，即使用最后使用的entry
*/
private void removeLast() {
//链表尾不为空,则将链表尾指向null. 删除连表尾（删除最少使用的缓存对象）
if (last != null) {
if (last.prev != null)
last.prev.next = null;
else
first = null;
last = last.prev;
}
}
/**
* 移动到链表头，表示这个节点是最新使用过的
*/
private void moveToHead(Entry node) {
if (node == first)
return;
if (node.prev != null)
node.prev.next = node.next;
if (node.next != null)
node.next.prev = node.prev;
if (last == node)
last = node.prev;
if (first != null) {
node.next = first;
first.prev = node;
}
first = node;
node.prev = null;
if (last == null)
last = first;
}
/*
* 清空缓存
*/
public void clear() {
first = null;
last = null;
currentSize = 0;
}
}
class Entry {
Entry prev;//前一节点
Entry next;//后一节点
Object value;//值
Object key;//键
}

public class LRUCache {
	
	private int cacheSize;
	private Hashtable<Object, Entry> nodes;//缓存容器
	private int currentSize;
	private Entry first;//链表头
	private Entry last;//链表尾
	
	public LRUCache(int i) {
		currentSize = 0;
		cacheSize = i;
		nodes = new Hashtable<Object, Entry>(i);//缓存容器
	}
	
	/**
	 * 获取缓存中对象,并把它放在最前面
	 */
	public Entry get(Object key) {
		Entry node = nodes.get(key);
		if (node != null) {
			moveToHead(node);
			return node;
		} else {
			return null;
		}
	}
	
	/**
	 * 添加 entry到hashtable, 并把entry 
	 */
	public void put(Object key, Object value) {
		//先查看hashtable是否存在该entry, 如果存在，则只更新其value
		Entry node = nodes.get(key);
		
		if (node == null) {
			//缓存容器是否已经超过大小.
			if (currentSize >= cacheSize) {
				nodes.remove(last.key);
				removeLast();
			} else {
				currentSize++;
			}			
			node = new Entry();
		}
		node.value = value;
		//将最新使用的节点放到链表头，表示最新使用的.
		moveToHead(node);
		nodes.put(key, node);
	}

	/**
	 * 将entry删除, 注意：删除操作只有在cache满了才会被执行
	 */
	public void remove(Object key) {
		Entry node = nodes.get(key);
		//在链表中删除
		if (node != null) {
			if (node.prev != null) {
				node.prev.next = node.next;
			}
			if (node.next != null) {
				node.next.prev = node.prev;
			}
			if (last == node)
				last = node.prev;
			if (first == node)
				first = node.next;
		}
		//在hashtable中删除
		nodes.remove(key);
	}

	/**
	 * 删除链表尾部节点，即使用最后 使用的entry
	 */
	private void removeLast() {
		//链表尾不为空,则将链表尾指向null. 删除连表尾（删除最少使用的缓存对象）
		if (last != null) {
			if (last.prev != null)
				last.prev.next = null;
			else
				first = null;
			last = last.prev;
		}
	}
	
	/**
	 * 移动到链表头，表示这个节点是最新使用过的
	 */
	private void moveToHead(Entry node) {
		if (node == first)
			return;
		if (node.prev != null)
			node.prev.next = node.next;
		if (node.next != null)
			node.next.prev = node.prev;
		if (last == node)
			last = node.prev;
		if (first != null) {
			node.next = first;
			first.prev = node;
		}
		first = node;
		node.prev = null;
		if (last == null)
			last = first;
	}
	/*
	 * 清空缓存
	 */
	public void clear() {
		first = null;
		last = null;
		currentSize = 0;
	}

}

class Entry {
	Entry prev;//前一节点
	Entry next;//后一节点
	Object value;//值
	Object key;//键
}

转载请注明出处：http://blog.csdn.net/beiyeqingteng

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