package com.blade.kit;
import java.util.*;
/**
* @author darren
* @date 2019/10/16 13:39
*/
public class LRUSet<E> extends AbstractSet<E>{
private transient HashMap<E, Object> map;
private static final Object PRESENT = new Object();
public LRUSet(int capacity) {
this.map = new LinkedHashMap<E, Object>(Math.min(32,capacity), .75f, true) {
@Override
protected boolean removeEldestEntry(Map.Entry eldest) {
// 这个capacity,准确说应该是initialCapacity,值不会改变了。理论上这个LRUSet最多扩容一次
return size() > capacity;
}
};
}
@Override
public int size() {
return map.size();
}
@Override
public boolean isEmpty() {
return map.isEmpty();
}
@Override
public boolean contains(Object o) {
// 只要map的value存在值就是不为空
return map.get(o) == PRESENT;
}
@Override
public Iterator<E> iterator() {
return map.keySet().iterator();
}
@Override
public boolean add(E e) {
// map的put方法会返回原来的value,如果原来为null那么就添加成功,如果原来已经有了value,那么就是false。所有的put,也都是放一个空对象Object
return map.put(e, PRESENT) == null;
}
@Override
public boolean remove(Object o) {
return map.remove(o) == PRESENT;
}
@Override
public void clear() {
map.clear();
}
}
从作者描述的观点来看,修复确实很严谨。后续使用java collection的时候都应该注意这些边界值。极大或者极小值得处理。
本次修复的重点看LinkedHashMap
中的removeEldestEntry()
这个方法
/**
* 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;
}
利用上述这个方法,完成了一个LRU的实现。
LinkedHashMap
不需要扩容(这个提问有误),那么为啥需要loadFactor负载因子
呢?
答:LinkedHashMap
是继承HashMap的,底层的逻辑是一样的,只是增加了前后的对象存储
,所以也需要扩容
,此外应该上面LRUSet的指定了loadFactor
,为了满足HashMap的需求。
/**
* Constructs an empty <tt>LinkedHashMap</tt> instance with the
* specified initial capacity, load factor and ordering mode.
*
* @param initialCapacity the initial capacity
* @param loadFactor the load factor
* @param accessOrder the ordering mode - <tt>true</tt> for
* access-order, <tt>false</tt> for insertion-order
* @throws IllegalArgumentException if the initial capacity is negative
* or the load factor is nonpositive
*/
public LinkedHashMap(int initialCapacity,
float loadFactor,
boolean accessOrder) {
super(initialCapacity, loadFactor);
this.accessOrder = accessOrder;
}
不常见的一种for循环使用方式
public boolean containsValue(Object value) {
for (LinkedHashMap.Entry<K,V> e = head; e != null; e = e.after) {
V v = e.value;
if (v == value || (value != null && value.equals(v)))
return true;
}
return false;
}
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