• LinkedHashMap 和 LRU算法实现


    个人觉得LinkedHashMap 存在的意义就是为了实现 LRU 算法。

    public class LinkedHashMap<K,V> extends HashMap<K,V>
        implements Map<K,V>
    {
        public LinkedHashMap(int initialCapacity,
                             float loadFactor,
                             boolean accessOrder) {
            super(initialCapacity, loadFactor);
            this.accessOrder = accessOrder;
        }
        ....

    1、LinkedHashMap 的 <K,V>用HashMap存储。

    2、LinkedHashMap 的Key 用双向链表维护。

      当用get 和 set 方法的时候,内部维护key的双向链表的结构顺序会变动。

    3、accessOrder:false 基于插入顺序  true  基于访问顺序(get一个元素后,这个元素被加到最后,使用了LRU  最近最少被使用的调度算法)。

    4、removeEldestEntry方法,考虑清楚是否要重载。如果最大容量固定,则需要重载,否则表现为自适应

    protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
            return false;
        }

    最简单的LRU算法实现

    update1:第二个实现,读操作不必要采用独占锁,缓存显然是读多于写,读的时候一开始用独占锁是考虑到要递增计数和更新时间戳要加锁,不过这两个变量都是采用原子变量,因此也不必采用独占锁,修改为读写锁。
    update2:一个错误,老是写错关键字啊,LRUCache的maxCapacity应该声明为volatile,而不是transient。
       
       最简单的LRU算法实现,就是利用jdk的LinkedHashMap,覆写其中的removeEldestEntry(Map.Entry)方法即可,如下所示:

    import java.util.ArrayList;
    import java.util.Collection;
    import java.util.LinkedHashMap;
    import java.util.concurrent.locks.Lock;
    import java.util.concurrent.locks.ReentrantLock;
    import java.util.Map;
    
    
    /**
     * 类说明:利用LinkedHashMap实现简单的缓存, 必须实现removeEldestEntry方法,具体参见JDK文档
     * 
     * @author dennis
     * 
     * @param <K>
     * @param <V>
     */
    public class LRULinkedHashMap<K, V> extends LinkedHashMap<K, V> {
        private final int maxCapacity;
    
        private static final float DEFAULT_LOAD_FACTOR = 0.75f;
    
        private final Lock lock = new ReentrantLock();
    
        public LRULinkedHashMap(int maxCapacity) {
            super(maxCapacity, DEFAULT_LOAD_FACTOR, true);
            this.maxCapacity = maxCapacity;
        }
    
        @Override
        protected boolean removeEldestEntry(java.util.Map.Entry<K, V> eldest) {
            return size() > maxCapacity;
        }
        @Override
        public boolean containsKey(Object key) {
            try {
                lock.lock();
                return super.containsKey(key);
            } finally {
                lock.unlock();
            }
        }
    
        
        @Override
        public V get(Object key) {
            try {
                lock.lock();
                return super.get(key);
            } finally {
                lock.unlock();
            }
        }
    
        @Override
        public V put(K key, V value) {
            try {
                lock.lock();
                return super.put(key, value);
            } finally {
                lock.unlock();
            }
        }
    
        public int size() {
            try {
                lock.lock();
                return super.size();
            } finally {
                lock.unlock();
            }
        }
    
        public void clear() {
            try {
                lock.lock();
                super.clear();
            } finally {
                lock.unlock();
            }
        }
    
        public Collection<Map.Entry<K, V>> getAll() {
            try {
                lock.lock();
                return new ArrayList<Map.Entry<K, V>>(super.entrySet());
            } finally {
                lock.unlock();
            }
        }
    }

        如果你去看LinkedHashMap的源码可知,LRU算法是通过双向链表来实现,当某个位置被命中,通过调整链表的指向将该位置调整到头位置,新加入的内容直接放在链表头,如此一来,最近被命中的内容就向链表头移动,需要替换时,链表最后的位置就是最近最少使用的位置。
        LRU算法还可以通过计数来实现,缓存存储的位置附带一个计数器,当命中时将计数器加1,替换时就查找计数最小的位置并替换,结合访问时间戳来实现。这种算法比较适合缓存数据量较小的场景,显然,遍历查找计数最小位置的时间复杂度为O(n)。我实现了一个,结合了访问时间戳,当最小计数大于MINI_ACESS时(这个参数的调整对命中率有较大影响),就移除最久没有被访问的项:

    package net.rubyeye.codelib.util.concurrency.cache;
    
    import java.io.Serializable;
    import java.util.ArrayList;
    import java.util.Collection;
    import java.util.HashMap;
    import java.util.Iterator;
    import java.util.Map;
    import java.util.Set;
    import java.util.concurrent.atomic.AtomicInteger;
    import java.util.concurrent.atomic.AtomicLong;
    import java.util.concurrent.locks.Lock;
    import java.util.concurrent.locks.ReadWriteLock;
    import java.util.concurrent.locks.ReentrantLock;
    import java.util.concurrent.locks.ReentrantReadWriteLock;
    
    /**
     * 
     * @author dennis 类说明:当缓存数目不多时,才用缓存计数的传统LRU算法
     * @param <K>
     * @param <V>
     */
    public class LRUCache<K, V> implements Serializable {
    
        private static final int DEFAULT_CAPACITY = 100;
    
        protected Map<K, ValueEntry> map;
    
        private final ReadWriteLock lock = new ReentrantReadWriteLock();
    
        private final Lock readLock = lock.readLock();
    
        private final Lock writeLock = lock.writeLock();
    
        private final volatile int maxCapacity;  //保持可见性
    
        public static int MINI_ACCESS = 5;
    
        public LRUCache() {
            this(DEFAULT_CAPACITY);
        }
    
        public LRUCache(int capacity) {
            if (capacity <= 0)
                throw new RuntimeException("缓存容量不得小于0");
            this.maxCapacity = capacity;
            this.map = new HashMap<K, ValueEntry>(maxCapacity);
        }
    
        public boolean ContainsKey(K key) {
            try {
                readLock.lock();
                return this.map.containsKey(key);
            } finally {
                readLock.unlock();
            }
        }
    
        public V put(K key, V value) {
            try {
                writeLock.lock();
                if ((map.size() > maxCapacity - 1) && !map.containsKey(key)) {
                    // System.out.println("开始");
                    Set<Map.Entry<K, ValueEntry>> entries = this.map.entrySet();
                    removeRencentlyLeastAccess(entries);
                }
                ValueEntry new_value = new ValueEntry(value);
                ValueEntry old_value = map.put(key, new_value);
                if (old_value != null) {
                    new_value.count = old_value.count;
                    return old_value.value;
                } else
                    return null;
            } finally {
                writeLock.unlock();
            }
        }
    
        /**
         * 移除最近最少访问
         */
        protected void removeRencentlyLeastAccess(
                Set<Map.Entry<K, ValueEntry>> entries) {
            // 最小使用次数
            long least = 0;
            // 访问时间最早
            long earliest = 0;
            K toBeRemovedByCount = null;
            K toBeRemovedByTime = null;
            Iterator<Map.Entry<K, ValueEntry>> it = entries.iterator();
            if (it.hasNext()) {
                Map.Entry<K, ValueEntry> valueEntry = it.next();
                least = valueEntry.getValue().count.get();
                toBeRemovedByCount = valueEntry.getKey();
                earliest = valueEntry.getValue().lastAccess.get();
                toBeRemovedByTime = valueEntry.getKey();
            }
            while (it.hasNext()) {
                Map.Entry<K, ValueEntry> valueEntry = it.next();
                if (valueEntry.getValue().count.get() < least) {
                    least = valueEntry.getValue().count.get();
                    toBeRemovedByCount = valueEntry.getKey();
                }
                if (valueEntry.getValue().lastAccess.get() < earliest) {
                    earliest = valueEntry.getValue().count.get();
                    toBeRemovedByTime = valueEntry.getKey();
                }
            }
            // System.out.println("remove:" + toBeRemoved);
            // 如果最少使用次数大于MINI_ACCESS,那么移除访问时间最早的项(也就是最久没有被访问的项)
            if (least > MINI_ACCESS) {
                map.remove(toBeRemovedByTime);
            } else {
                map.remove(toBeRemovedByCount);
            }
        }
    
        public V get(K key) {
            try {
                readLock.lock();
                V value = null;
                ValueEntry valueEntry = map.get(key);
                if (valueEntry != null) {
                    // 更新访问时间戳
                    valueEntry.updateLastAccess();
                    // 更新访问次数
                    valueEntry.count.incrementAndGet();
                    value = valueEntry.value;
                }
                return value;
            } finally {
                readLock.unlock();
            }
        }
    
        public void clear() {
            try {
                writeLock.lock();
                map.clear();
            } finally {
                writeLock.unlock();
            }
        }
    
        public int size() {
            try {
                readLock.lock();
                return map.size();
            } finally {
                readLock.unlock();
            }
        }
    
        public long getCount(K key) {
            try {
                readLock.lock();
                ValueEntry valueEntry = map.get(key);
                if (valueEntry != null) {
                    return valueEntry.count.get();
                }
                return 0;
            } finally {
                readLock.unlock();
            }
        }
    
        public Collection<Map.Entry<K, V>> getAll() {
            try {
                readLock.lock();
                Set<K> keys = map.keySet();
                Map<K, V> tmp = new HashMap<K, V>();
                for (K key : keys) {
                    tmp.put(key, map.get(key).value);
                }
                return new ArrayList<Map.Entry<K, V>>(tmp.entrySet());
            } finally {
                readLock.unlock();
            }
        }
    
        class ValueEntry implements Serializable {
            private V value;
    
            private AtomicLong count;
    
            private AtomicLong lastAccess;
    
            public ValueEntry(V value) {
                this.value = value;
                this.count = new AtomicLong(0);
                lastAccess = new AtomicLong(System.nanoTime());
            }
    
            public void updateLastAccess() {
                this.lastAccess.set(System.nanoTime());
            }
    
        }
    }

    参考:

    简单LRU算法实现缓存-update2

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  • 原文地址:https://www.cnblogs.com/549294286/p/3855648.html
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