[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 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.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 size
Node 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 tail
int 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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