【leetcode刷题笔记】LRU Cache

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.

1. get(key)：如果链表中有key对应的node，返回该node的值，并把node移到链表头部（最近使用过）；如果没有，返回-1；
2. set(key,value)：如果链表中已经有key对应的node，修改对应node的值为value（但不需要把这个node放在头结点处）；如果链表中没有key对应的node，那么要新建node插入链表中，但此时要看链表是否还有空间。如果没有，就将尾部node（最少使用的node）删除，然后在头部插入新的node。

为了提高查找效率，这里使用一个hashMap存放<key,node>键值对，这样就可以在O(1)的时间判断cache中是否存在对应的key。

代码如下：

 1 public class LRUCache {
 2     private class Node{
 3         int value;
 4         int key;
 5         Node before;
 6         Node after;
 7         public Node(int key,int value){
 8             this.value = value;
 9             this.key = key;
10             before = null;
11             after = null;
12         }
13     }
14
15     private int capacity;
16     private HashMap<Integer, Node> map = new HashMap<Integer,Node>();
17     private Node headNode = new Node(-1, -1);
18     private Node tailNode = new Node(-1, -1);
19
20     public LRUCache(int capacity) {
21         this.capacity = capacity;
22         headNode.after = tailNode;
23         tailNode.before = headNode;
24     }
25
26     public void move_to_head(Node current){
27         current.after = headNode.after;
28         headNode.after = current;
29         current.before = headNode;
30         current.after.before = current;
31     }
32     public int get(int key) {
33         //if we don‘t have this node
34         if(!map.containsKey(key))
35             return -1;
36
37         //if we have this node, get it and move it to head
38         Node current = map.get(key);
39         current.before.after = current.after;
40         current.after.before = current.before;
41         move_to_head(current);
42
43         return map.get(key).value;
44     }
45
46     public void set(int key, int value) {
47         //if we already have this node,just change its value
48         if(get(key) != -1){
49             map.get(key).value = value;
50             return;
51         }
52
53         //if we indeed don‘t have this node, we first check capacity
54         if(map.size() == capacity){
55             map.remove(tailNode.before.key);
56             tailNode.before.before.after = tailNode;
57             tailNode.before = tailNode.before.before;
58         }
59
60         //now we are sure we have space for this new node,put it ahead of the list
61         Node newNode = new Node(key, value);
62         map.put(key, newNode);
63         move_to_head(newNode);
64     }
65 }

在实现的时候，还设置了两个node：head和tail，真正的cache数据节点存放在二者之间。

【leetcode刷题笔记】LRU Cache