public class RetainBestCache<K, T extends Rankable> {
private Map<K, T> cache;. 鍥磋鎴戜滑@1point 3 acres
private Map<Long, Set<K>> rankingOfObject;.鏈枃鍘熷垱鑷�1point3acres璁哄潧
private DataSource<K, T> dataSource;
private int maxSizeOfCache;
/* Constructor with a data source (assumed to be slow) and a cache size */
public RetainBestCache(DataSource<K,T> ds, int entriesToRetain) {. From 1point 3acres bbs
// Implementation here
cache = new HashMap<>();
rankingOfObject = new TreeMap<>();
dataSource = ds;
maxSizeOfCache = entriesToRetain;
}
/* Gets some data. If possible, retrieves it from cache to be fast. If the data is not cached,
* retrieves it from the data source. If the cache is full, attempt to cache the returned data,
* evicting the T with lowest rank among the ones that it has available
* If there is a tie, the cache may choose any T with lowest rank to evict.
*/
public T get(K key) {
// Implementation here
if(cache.containsKey(key)) {
return cache.get(key);
}
return fetchDataFromDs(key);
}
. visit 1point3acres.com for more.
private T fetchDataFromDs(K key) {
if(cache.size() >= maxSizeOfCache) {
evictElement();
}
T object = dataSource.get(key);
cache.put(key, object);
long rankOfObject = object.getRank();
if(!rankingOfObject.containsKey(rankOfObject)) {. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
rankingOfObject.put(rankOfObject, new HashSet<>());
}
rankingOfObject.get(rankOfObject).add(key);
return object;
}
private void evictElement() {
Entry<Long, Set<K>> entry = rankingOfObject.topEntry();
K key = entry.getValue().getIterator().next();
entry.getValue().remove(key);
cache.remove(key)
if(entry.getValue().size() == 0) {
rankingOfObject.remove(entry.getKey());
}.鏈枃鍘熷垱鑷�1point3acres璁哄潧
}
}
. visit 1point3acres.com for more.
// What if rank is defined as number of reads of element in cache?
// LRU. from: 1point3acres.com/bbs
// Let's assume that rank can change dynamicly. It is not immutable, but it is not LRU. We do not know how it is changed
/*
* For reference, here are the Rankable and DataSource interfaces.
* You do not need to implement them, and should not make assumptions
* about their implementations.
*/
public interface Rankable {. from: 1point3acres.com/bbs
/**.鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
* Returns the Rank of this object, using some algorithm and potentially
* the internal state of the Rankable.
*/
long getRank();
}
public interface DataSource<K, T extends Rankable> {
T get (K key);
}. W
