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电面介绍自己之后是两道题，一个是leetcode上 Nested list weight sum 那道，我做出来了。

但是第二道是一道cach

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现它不是, 这是题目是这样的，网上有答案

public class RetainBestCache<K, T extends Rankable> {
int entriesToRetain;
HashMap<K, T> map = new HashMap<K,T>();
DataSource<K,T> ds;
/* Constructor with a data source (assumed to be slow) and a cache size */
public RetainBestCache(DataSource<K,T> ds, int entriesToRetain) {
//impliment here
}
/* 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) {
//impliment here
}
/*
* 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 {
/**
* 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);
}

复制代码

密码

[quote]liqingfd 发表于 2019-5-3 11:12
刚刚写了一个版本，跟楼上grandyang链接里面的差不多，欢迎指正!

import java.ut ...[/quote]
[hide=200]没必要用TreeMap吧，感觉PriorityQueue应该就够了吧？[code]public class RetainBestCache<K, T extends Rankable> {
int entriesToRetain;
HashMap<K, T> map = new HashMap<K,T>();
PriorityQueue<Wrapper<K, T>> pq;
DataSource<K,T> ds;
/* Constructor with a data source (assumed to be slow) and a cache size */
public RetainBestCache(DataSource<K,T> ds, int entriesToRetain) {
//impliment here
this.pq = new PriorityQueue<>(new Comparator<Wrapper>() {
public int compare(Wrapper w1, Wrapper w2) {
return w1.data.getRank() - w2.data.getRank();
}
});
this.ds = ds;
this.entriesToRetain = 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) {
//impliment here
if (map.containsKey(key)) {
return map.get(key);
}
T data = DataSource.get(key);
if (map.size() < entriesToRetain) {
map.put(key, data);
pq.offer(new Wrapper(key, data));
} else {
evict();
map.put(key, data);
pq.offer(new Wrapper(key, data));
}
return data;
}
private evict() {
Wrapper leastRank = pq.poll();
map.remove(leastRank.key);
}
}
class Wrapper<K, T> {
T data;
K key;
public Wrapper(T data, K key) {
this.data = data;
this.key = key;
}
}
/*
* 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 {
/**
* 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);
}

复制代码

[/hide]

Miaaaaa

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Wyf2222

请问楼主怎么做的
Cache hit o(1) , cache miss tree map ologn 的方案面试官接受吗

jeh

Wyf2222 发表于 2019-2-22 09:51
请问楼主怎么做的
Cache hit o(1) , cache miss tree map ologn 的方案面试官接受吗

我没有做出来来，还没说到那个程度，我只找到这个网上的答案： https://www.cnblogs.com/apanda009/p/7945036.html

jeh

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Wu_kong

请问楼主是LRU 或者 LFU吗？

jeh

gui59106375 发表于 2019-2-22 11:44
请问楼主是LRU 或者 LFU吗？

不是的，我上面的链接给了网上的解法

Ellachen

用PriorityQueue也可以

WantBetterLife

谢谢楼主cache的分享

huntersjm

cache 这题很有意思，用最小堆实现吗？超过size把堆顶remove掉，把新的add进去

liqingfd

刚刚写了一个版本，跟楼上grandyang链接里面的差不多，欢迎指正!

import java.util.*;
public class RetainBestCache<K, T extends RetainBestCache.Rankable> {
int entriesToRetain;
Map<K, T> keyValueMap; // map<key, val>
Map<Long, List<K>> rankKeyMap; // map<rank, list<key>>
DataSource<K, T> ds;
/* Constructor with a data source (assumed to be slow) and a cache size */
public RetainBestCache(DataSource<K, T> ds, int entriesToRetain) {
this.keyValueMap = new HashMap<>();
this.rankKeyMap = new TreeMap<>();
this.ds = ds;
this.entriesToRetain = 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) {
if (keyValueMap.containsKey(key)) {
return keyValueMap.get(key);
}
T val = ds.get(key);
keyValueMap.put(key, val);
rankKeyMap.put(val.getRank(), new ArrayList<>());
rankKeyMap.get(val.getRank()).add(key);
while (keyValueMap.size() > entriesToRetain) {
Long lowestRank = rankKeyMap.keySet().iterator().next();
K k = rankKeyMap.get(lowestRank).get(0);
rankKeyMap.get(lowestRank).remove(0);
if (rankKeyMap.get(lowestRank).size() == 0) {
rankKeyMap.remove(lowestRank);
}
keyValueMap.remove(k);
}
return val;
}
/*
* 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 {
/**
* 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);
}
}

复制代码

补充内容 (2019-5-5 06:04):
line 28写错了，应该先判断一下map里面有没有这个key再put的。

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