设计一个带有过期时间的LRU缓存。
仍旧是同事面试时遇到的题，记录一下我的解法。
经典的LRU题目，leetcode上有初始版本，lru-cache。
这道面试题在leetcode题目的基础上增加了过期时间，到达容量上限后，先逐出过期的数据，然后再逐出最久未使用的数据。
仿照leetcode的题目，本题大致是实现这样一个数结构。
class LRUCache {
public LRUCache(int capacity) {
}
public int get(int key) {
}
public void put(int key, int value, long expire) {
}
}
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value, expire);
*/
LinkedHashMap + 惰性删除
众所周知，双向链表 + hash map可以实现LRU，而Java里LinkedHashMap正好两者都有。可以说，用LinkedHashMap实现LRU，也是众多Javaer面试必会的题目之一。
本题还需要处理数据过期，而逐出过期数据不由让人想起Redis的惰性删除策略。
利用LinkedHashMap，再仿照Redis的惰性删除
public class LRUCache extends LinkedHashMap<Integer, LRUCache.Node> {
int capacity;
public LRUCache(int capacity) {
super(capacity, 0.75f, true);
this.capacity = capacity;
}
public int get(int key) {
Node node = super.get(key);
if (node == null) {
return -1;
}
if (node.expire < System.currentTimeMillis()) {
remove(node);
return -1;
}
return node.val;
}
public void put(int key, int value, long expire) {
Node node = new Node(key, value, expire);
super.put(key, node);
}
@Override
protected boolean removeEldestEntry(Map.Entry<Integer, Node> eldest) {
if (size() <= capacity) {
return false;
}
long now = System.currentTimeMillis();
if (eldest.getValue().expire < now) {
return true;
}
Iterator<Node> iterator = values().iterator();
while (iterator.hasNext()) {
Node node = iterator.next();
if (node.expire < now) {
iterator.remove();
return false;
}
}
return true;
}
class Node {
private int key;
private int val;
private long expire;
public Node(int key, int val, long expire) {
this.key = key;
this.val = val;
this.expire = expire;
}
}
}
再仔细看看LinkedHashMap的get和put方法，了解下具体过程。
get过程
java.util.LinkedHashMap#get在获取数据之后，如果按访问排序（accessOrder=true)，会执行afterNodeAccess方法，将数据移到双向链表尾部。
/**
* The iteration ordering method for this linked hash map: {@code true}
* for access-order, {@code false} for insertion-order.
*
* @serial
*/
final boolean accessOrder;
/**
* Returns the value to which the specified key is mapped,
* or {@code null} if this map contains no mapping for the key.
*
* <p>More formally, if this map contains a mapping from a key
* {@code k} to a value {@code v} such that {@code (key==null ? k==null :
* key.equals(k))}, then this method returns {@code v}; otherwise
* it returns {@code null}. (There can be at most one such mapping.)
*
* <p>A return value of {@code null} does not <i>necessarily</i>
* indicate that the map contains no mapping for the key; it's also
* possible that the map explicitly maps the key to {@code null}.
* The {@link #containsKey containsKey} operation may be used to
* distinguish these two cases.
*/
public V get(Object key) {
Node<K,V> e;
if ((e = getNode(key)) == null)
return null;
if (accessOrder)
afterNodeAccess(e);
return e.value;
}
void afterNodeAccess(Node<K,V> e) { // move node to last
LinkedHashMap.Entry<K,V> last;
if (accessOrder && (last = tail) != e) {
LinkedHashMap.Entry<K,V> p =
(LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
p.after = null;
if (b == null)
head = a;
else
b.after = a;
if (a != null)
a.before = b;
else
last = b;
if (last == null)
head = p;
else {
p.before = last;
last.after = p;
}
tail = p;
++modCount;
}
}
put过程
put的过程比get更复杂一些，首先是java.util.HashMap#put和putVal，这代码也是经典Java面试题目之一，不同jdk版本实现也不一样，截取jdk 17👇
/**
* Associates the specified value with the specified key in this map.
* If the map previously contained a mapping for the key, the old
* value is replaced.
*
* @param key key with which the specified value is to be associated
* @param value value to be associated with the specified key
* @return the previous value associated with {@code key}, or
* {@code null} if there was no mapping for {@code key}.
* (A {@code null} return can also indicate that the map
* previously associated {@code null} with {@code key}.)
*/
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
/**
* Implements Map.put and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to put
* @param onlyIfAbsent if true, don't change existing value
* @param evict if false, the table is in creation mode.
* @return previous value, or null if none
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
跟本题相关的重点是afterNodeAccess和afterNodeInsertion方法。afterNodeAccess在get时也会调用，将数据移到双向链表尾部。
void afterNodeInsertion(boolean evict) { // possibly remove eldest
LinkedHashMap.Entry<K,V> first;
if (evict && (first = head) != null && removeEldestEntry(first)) {
K key = first.key;
removeNode(hash(key), key, null, false, true);
}
}
protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
return false;
}
put操作的最后一步是调用afterNodeInsertion，通过removeEldestEntry判断是否需要删除元素。
本题的解答就是重写了removeEldestEntry，来实现必要时删除过期元素。
其他
总的来说，过期key删除有三个策略：
- 定时删除：创建定时器，到达过期时间时，就删除key。
- 最简单的实现：new一个线程，死循环执行删除任务。
- 内存友好，不会让过期数据占存储空间，但是会增加CPU压力。
- 惰性删除：到必要时候才删除key，比如get一个过期key时。
- 内存不友好，CPU友好
- 定期删除：间隔一段时间，随机抽样一批key，删除其中过期的
- 定时删除和惰性删除的调和版本，可以通过调参来控制对CPU和内存影响。
本文只提供了惰性删除策略的实现，网上也有很多实现定时删除的文章，可以参考。
此外，LinkedHashMap并不是线程安全的，如果需要保证线程安全，可以换成ConcurrentLinkedDeque + ConcurrentHashMap。
