[LeetCode]146. LRU Cache 深入浅出讲解和代码示例

1、汇总概要

以下思路涵盖了哈希与双向链表的结合使用、缓存设计等知识点

2、题目

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(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.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );cache.put(1, 1);cache.put(2, 2);cache.get(1);       // returns 1cache.put(3, 3);    // evicts key 2cache.get(2);       // returns -1 (not found)cache.put(4, 4);    // evicts key 1cache.get(1);       // returns -1 (not found)cache.get(3);       // returns 3cache.get(4);       // returns 4

3、审题

设计一个简单版的最近使用缓存模型。缓存空间有容量限制，时间复杂度要求是O(1)。

其中“最近使用”是指最近被访问过(被cache.get调用过）。

4、解题思路

以上对cache的操作有：添加(put)、查找(get)、替换(put)，因有容量限制，还需有删除，每次当容量满时，将最久未使用的节点删除。

为快速添加和删除，我们可以用双向链表来设计cache，链表中从头到尾的数据顺序依次是，(最近访问)->...(最旧访问)：

1）添加节点：新节点插入到表头即可，时间复杂度O(1)；

2）查找节点：每次节点被查询到时，将节点移动到链表头部，时间复杂度O(n)

3)  替换节点：查找到后替换(更新节点value)，将节点移动到链表头部；

可见在查找节点时，因对链表需遍历，时间复杂度O(n)，为达到O(1)，可以考虑数据结构中加入哈希(hash)。

=>我们需要用两种数据结构来解题：双向链表、哈希表

示意图如下：

5、代码示例 - Java

import java.util.*;class Node{int key;int value;Node next;Node pre;public Node(int key,int value,Node pre, Node next){this.key = key;this.value = value;this.pre = pre;this.next = next;}}public class LRUCache {int capacity;int count;//cache sizeNode head;Node tail;HashMap<Integer,Node>hm;    public LRUCache(int capacity) { //only initial 2 Node is enough, head/tail    this.capacity = capacity;    this.count = 2;    this.head = new Node(-1,-1,null,null);    this.tail = new Node(-2,-2,this.head,null);    this.head.next = this.tail;        hm = new HashMap<Integer,Node>();        hm.put(this.head.key, this.head);        hm.put(this.tail.key, this.tail);    }        public int get(int key) {    int value = -1;    if(hm.containsKey(key)){    Node nd = hm.get(key);    value = nd.value;    detachNode(nd); //detach nd from current place    insertToHead(nd); //insert nd into head    }return value;    }        public void put(int key, int value) {    if(hm.containsKey(key)){ //update    Node nd = hm.get(key);    nd.value = value;    //move to head    detachNode(nd); //detach nd from current place    insertToHead(nd); //insert nd into head    }else{ //add    Node newNd = new Node(key,value,null,this.head);    this.head.pre = newNd; //insert into head    this.head = newNd;    hm.put(key, newNd); //add into hashMap    this.count ++;    if(this.count > capacity){ //need delete node    removeNode();    }    }    }    //common func    public void insertToHead(Node nd){    this.head.pre = nd;    nd.next = this.head;    nd.pre = null;    this.head = nd;    }    public void detachNode(Node nd){    nd.pre.next = nd.next;    if(nd.next!=null){    nd.next.pre = nd.pre;    }else{    this.tail = nd.pre;    }    }    public void removeNode(){ //remove from tailint tailKey = this.tail.key;this.tail = this.tail.pre;this.tail.next = null;hm.remove(tailKey);this.count --;    }    public void printCache(){    System.out.println("\nPRINT CACHE ------ ");    System.out.println("count: "+count);    System.out.println("From head:");    Node p = this.head;    while(p!=null){    System.out.println("key: "+p.key+" value: "+p.value);    p = p.next;    }    System.out.println("From tail:");    p = this.tail;    while(p!=null){    System.out.println("key: "+p.key+" value: "+p.value);    p = p.pre;    }        }        public static void main(String[] args){    LRUCache lc = new LRUCache(3);    lc.printCache();        lc.put(1, 1);    lc.put(2, 2);    lc.put(3, 3);    lc.printCache();        lc.get(2);    lc.printCache();        lc.put(4, 4);    lc.printCache();        lc.get(1);    lc.printCache();        lc.put(3, 33);    lc.printCache();    }}

【注：】这里要区分下hashmap和hashtable在java中使用的区别(继承于不同的类、线程安全、扩容等方面)

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