LRU Cache的实现
2016-03-25 15:55
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LeetCode 146 LRU Cache
我的LeetCode题解GitHub链接https://github.com/YuanSun927/LeetCode,欢迎交流!
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
操作规则如下:
1. 新加入的Node插入链表头
2. 访问(取或更新)了已存在Node,则将该Node移到链表头
3. 若Map的size大于Cache容量,则删除链表尾节点
下面是LeetCode LRU Cache的AC代码。
以下代码不是LeetCode的解题代码,该实现利用了泛型,更通用。
我的LeetCode题解GitHub链接https://github.com/YuanSun927/LeetCode,欢迎交流!
1. 题目描述
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
2. 基于HashMap实现
通过HashMap建立Key与Value的映射。Value为Node类型,Node不仅包含key与value,还有一个指向上一个被访问节点的引用和一个指向下一个被访问节点的引用,这样Node就建立了一个双向链表,维持了Node的访问顺序。操作规则如下:
1. 新加入的Node插入链表头
2. 访问(取或更新)了已存在Node,则将该Node移到链表头
3. 若Map的size大于Cache容量,则删除链表尾节点
下面是LeetCode LRU Cache的AC代码。
public class LRUCache { private class Node { int key; int value; Node prev; // 上一个被访问的节点 Node next; // 下一个被访问的节点 public Node(int key, int value) { this.key = key; this.value = value; prev = null; next = null; } } private Map<Integer, Node> cache = new HashMap<>(); private int capacity; private Node head = new Node(-1, -1); private Node tail = new Node(-1, -1); public LRUCache(int capacity) { this.capacity = capacity; head.next = tail; tail.prev = head; } public int get(int key) { Node node = cache.get(key); if (node == null) return -1; // 访问了该节点,所以将节点移到链表尾 moveToTail(node); return node.value; } public void set(int key, int value) { Node node = cache.get(key); if (node != null) { node.value = value; // 更新了该节点,所以将节点移到链表尾 moveToTail(node); return; } if (cache.size() == capacity) { cache.remove(head.next.key); head.next = head.next.next; head.next.prev = head; } Node insert = new Node(key, value); cache.put(key, insert); // 新插入了节点,将该节点添加到链表尾 addToTail(insert); } private void moveToTail(Node node) { // remove node node.prev.next = node.next; node.next.prev = node.prev; // add to tail node.prev = tail.prev; node.next = tail; node.prev.next = node; tail.prev = node; } private void addToTail(Node node) { node.prev = tail.prev; node.next = tail; node.prev.next = node; tail.prev = node; } }
3. 基于LinkedHashMap实现
要实现LRU Cache,其实是要实现如何维持Cache的访问次序。HashMap的元素迭代顺序是不确定的,而LinkedHashMap正可以维持元素迭代顺序(插入顺序或访问顺序)。LinkedHashMap可以按访问顺序维持元素迭代顺序正好符合LRU Cache需求。以下代码不是LeetCode的解题代码,该实现利用了泛型,更通用。
public class LRUCache<K, V> extends LinkedHashMap<K, V> { private static final long serialVersionUID = 1L; // 缓存的容量 private int capacity; LRUCache(int capacity){ // 调用LinkedHashMap的构造器 // 初始容量initialCapacity = 16 // 负载因子loadFactor = 0.75 // accessOrder = true,表示按访问顺序迭代 super(16, 0.75f, true); this.capacity=capacity; } /** * removeEldestEntry方法用于判断是否该删除最老元素(进入map最久或最久未被访问的元素) * 在LinkedHashMap中,该方法放回false,相当于条件一直不满足,不会删除最老元素 * 当Map为Cache时,该方法非常有用,当map.size() > capacity时,删除最久未被访问的元素 */ @Override public boolean removeEldestEntry(Map.Entry<K, V> eldest){ return size() > capacity; } }
4. Android LruCache
android.util包中的LruCache正是使用了LinkedHashMap实现了LRU Cache。与上述实现不同的是,android.util.LruCache使用了对象组合,而不是类继承。同时,android.util.LruCache在put和get等方法中使用了synchronized实现了线程同步,所以该类线程安全。public class LruCache<K, V> { private final LinkedHashMap<K, V> map; ... public final V get(K key) { if (key == null) { throw new NullPointerException("key == null"); } V mapValue; synchronized (this) { mapValue = map.get(key); if (mapValue != null) { hitCount++; return mapValue; } missCount++; } ... } public final V put(K key, V value) { if (key == null || value == null) { throw new NullPointerException("key == null || value == null"); } V previous; synchronized (this) { putCount++; size += safeSizeOf(key, value); previous = map.put(key, value); if (previous != null) { size -= safeSizeOf(key, previous); } } ... } public synchronized final int size() { return size; } }
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