LRU Cache

描述

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.

分析

为了使查找、插入和删除都有较高的性能,这题的关键是要使用一个双向链表和一个HashMap,因为:

  • HashMap保存每个节点的地址,可以基本保证在O(1)时间内查找节点
  • 双向链表能后在O(1)时间内添加和删除节点,单链表则不行

具体实现细节:

  • 越靠近链表头部,表示节点上次访问距离现在时间最短,尾部的节点表示最近访问最少
  • 访问节点时,如果节点存在,把该节点交换到链表头部,同时更新hash表中该节点的地址
  • 插入节点时,如果cache的size达到了上限capacity,则删除尾部节点,同时要在hash表中删除对应的项;新节点插入链表头部
LRU Cche
Figure: LRU Cche

代码

Java中也有双向链表LinkedList, 但是 LinkedList 封装的太深,没有能在O(1)时间内删除中间某个元素的API(C++的list有个splice(), O(1), 所以本题C++可以放心使用splice()),于是我们只能自己实现一个双向链表。

本题有的人直接用 LinkedHashMap ,代码更短,但这是一种偷懒做法,面试官一定会让你自己重新实现。

// LRU Cache
// 时间复杂度O(logn),空间复杂度O(n)
public class LRUCache {
    private int capacity;
    private final HashMap<Integer, Node> map;
    private Node head;
    private Node end;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        map = new HashMap<>();
    }

    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 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);
        }
    }

    private void remove(Node n){
        if (n.prev !=null) {
            n.prev.next = n.next;
        } else {
            head = n.next;
        }

        if (n.next != null) {
            n.next.prev = n.prev;
        } else {
            end = n.prev;
        }

    }

    private void setHead(Node n){
        n.next = head;
        n.prev = null;

        if (head!=null ) head.prev = n;

        head = n;

        if(end == null) end = head;
    }

    // doubly linked list
    static class Node {
        int key;
        int value;
        Node prev;
        Node next;

        public Node(int key, int value) {
            this.key = key;
            this.value = value;
        }
    }
}

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