【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.

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题解：用双向链表实现一个LRU cache，这里自定义越靠近链表头的node，在越近的时间使用过。支持两个操作——get和set：

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：

public class LRUCache {
    private class Node{
        int value;
        int key;
        Node before;
        Node after;
        public Node(int key,int value){
            this.value = value;
            this.key = key;
            before = null;
            after = null;
        }
    }
    
    private int capacity;
    private HashMap<Integer, Node> map = new HashMap<Integer,Node>();
    private Node headNode = new Node(-1, -1);
    private Node tailNode = new Node(-1, -1);
    
    public LRUCache(int capacity) {
        this.capacity = capacity;
        headNode.after = tailNode;
        tailNode.before = headNode;
    }
    
    public void move_to_head(Node current){
        current.after = headNode.after;
        headNode.after = current;
        current.before = headNode;
        current.after.before = current;
    }
    public int get(int key) {
        //if we don't have this node
        if(!map.containsKey(key))
            return -1;
        
        //if we have this node, get it and move it to head
        Node current = map.get(key);
        current.before.after = current.after;
        current.after.before = current.before;
        move_to_head(current);
        
        return map.get(key).value;        
    }
    
    public void set(int key, int value) {
        //if we already have this node,just change its value
        if(get(key) != -1){
            map.get(key).value = value;
            return;
        }
        
        //if we indeed don't have this node, we first check capacity
        if(map.size() == capacity){
            map.remove(tailNode.before.key);
            tailNode.before.before.after = tailNode;
            tailNode.before = tailNode.before.before;
        }
        
        //now we are sure we have space for this new node,put it ahead of the list
        Node newNode = new Node(key, value);
        map.put(key, newNode);
        move_to_head(newNode);
    }
}

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。