LinkedHashMap的用法
LinkedHashMap,底层实现是在HashMap的基础上,添加了双向链表,可以根据访问顺序进行遍历,从最少访问到最频繁访问的升序访问。DOC描述如下:
LinkedHashMap(int,float,boolean) create a linked hash map whose order of iteration is the order in which its entries were last accessed, from least-recently accessed to most-recently access-order . This kind of map is well-suited to building LRU caches.
普通的LinkedHashMap
public class LinkedHashMapTest { public static void main(String[] args) { //插入顺序 System.out.println("插入顺序"); accessByInsertOrder(); } /** * new LinkedHashMap<>(),与普通的HashMap相似,按键值哈希进行遍历 */ private static void accessByInsertOrder() { LinkedHashMap<String, Integer> map = new LinkedHashMap<>(); map.put("a", 1); map.put("b", 2); map.put("c", 3); map.put("a", 4); for(Entry<String, Integer> e : map.entrySet()) { System.out.println(e.getKey() + "-" +e.getValue()); } } }
输出:
插入顺序 a-4 b-2 c-3
按顺序访问进行遍历
public class LinkedHashMapTest { public static void main(String[] args) { //访问顺序 System.out.println("访问顺序"); accessByAccessOrder(); } /** * new LinkedHashMap<>(16, 0.75f, true),accessOrder=true,表示按访问顺序进行遍历 */ private static void accessByAccessOrder() { LinkedHashMap<String, Integer> map = new LinkedHashMap<>(16, 0.75f, true); map.put("a", 1); map.put("b", 2); map.put("c", 3); map.put("a", 4); for(Entry<String, Integer> e : map.entrySet()) { System.out.println(e.getKey() + "-" +e.getValue()); } } }
输出:最新访问的数据在后边
访问顺序 b-2 c-3 a-4
LRU缓存
同上一篇文章
public class LinkedHashMapTest { public static void main(String[] args) { //LRU缓存 System.out.println("LRU顺序"); LRUCache<String, Integer> cache = new LRUCache(); for(int i = 0; i < 10; i++) { cache.put("a"+i, i+1); } cache.get("a0"); cache.get("a1"); cache.put("b", 10); for(Entry<String, Integer> e : cache.entrySet()) { System.out.println(e.getKey() + "-" +e.getValue()); } } /** * 通过LinkedHashMap实现LRU缓存 * @author Administrator * * @param <K> * @param <V> */ static class LRUCache<K,V> extends LinkedHashMap<K,V>{ static final int maxSize = 10; LRUCache(){ super(16, 0.75f, true); } @Override protected boolean removeEldestEntry(Map.Entry<K,V> eldest) { return size() > maxSize; } } }
输出:最新访问的数据在后边,其中a2-3被移除缓存
LRU顺序 a3-4 a4-5 a5-6 a6-7 a7-8 a8-9 a9-10 a0-1 a1-2 b-10
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