LRUCache
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
Analysis:
首先,对于cache,如果希望有O(1)的查找复杂度,肯定要用hashmap来保存key和对象的映射。对于LRU而言,问题在于如何用O(1)解决cache entry的替换问题。
简单的说,cache的存储是一个链表的话,那么只要保证从头到尾的顺序就是cache从新到旧的顺序就好了,对于任何一个节点,如果被访问了,那么就将该节点移至头部。如果cache已满,那么就把尾部的删掉,从头部插入新节点。
所以,需要用到两个数据结构.
hashmap, 保存key和对象位置的映射
list,保存对象新旧程度的序列,双向链表为O(1)
class Node{
int key;
int value;
Node pre;
Node next;
public Node(int key, int value){
this.key = key;
this.value = value;
}
}
public class LRUCache {
int capacity;
HashMap<Integer, Node> map;
Node head=null;
Node end=null;
public LRUCache(int capacity) {
this.capacity = capacity;
map = new HashMap<Integer, Node>(capacity);
}
public int get(int key) {
if(map.containsKey(key)){
Node n = map.get(key);
remove(n);
setHead(n);
return n.value;
}
return -1;
}
public void remove(Node n){
if(n.pre!=null){
n.pre.next = n.next;
}else{
head = n.next;
}
if(n.next!=null){
n.next.pre = n.pre;
}else{
end = n.pre;
}
}
public void setHead(Node n){
n.next = head;
n.pre = null;
if(head!=null)
head.pre = n;
head = n;
if(end ==null)
end = head;
}
public void set(int key, int value) {
if(map.containsKey(key)){
Node old = map.get(key);
old.value = value;
remove(old);
setHead(old);
}else{
Node created = new Node(key, value);
if(map.size()>=capacity){
map.remove(end.key);
remove(end);
setHead(created);
}else{
setHead(created);
}
map.put(key, created);
}
}
}
Caffeine
Caffeine is a modern in-memory local cache for Java 8