Question

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.

Solution

My first thought is to use a linked list and a hashmap. But remove and add element for linked list is O(n) per step.

So, the final idea is to implement a bi-directional linked list.

details

For this question, I submitted over 10 times and finally got accepted.

There are many details we need to check.

1. When deleting a node, we should modify both prev and next attributes of its neighbors.

2. Every time when we add a new node, we should check whether it is the first node.

3. When input capacity is 1, we should deal with it separately.

 // Construct double list node
class Node {
public Node prev;
public Node next;
private int val;
private int key;
public Node(int key, int val) {
this.key = key;
this.val = val;
}
public void setValue(int val) {
this.val = val;
}
public int getKey() {
return key;
}
public int getValue() {
return val;
}
} public class LRUCache {
private int capacity;
private Map<Integer, Node> map;
private Node head;
private Node tail; public LRUCache(int capacity) {
this.capacity = capacity;
map = new HashMap<Integer, Node>();
} private void moveToHead(Node target) {
// Check whether target is already at head
if (target.prev == null)
return;
// Check whether target is at tail
if (target == tail)
tail = target.prev;
Node prev = target.prev;
Node next = target.next;
if (prev != null)
prev.next = next;
if (next != null)
next.prev = prev; Node oldHead = head;
target.prev = null;
target.next = oldHead;
oldHead.prev = target;
head = target;
} public int get(int key) {
if (!map.containsKey(key))
return -1;
Node current = map.get(key);
// Move found node to head
moveToHead(current);
return current.getValue();
} public void set(int key, int value) {
if (map.containsKey(key)) {
Node current = map.get(key);
current.setValue(value);
// Move found node to head
moveToHead(current); } else {
Node current = new Node(key, value);
// Add new node to map
map.put(key, current); // Check whether map size is bigger than capacity
if (map.size() > capacity) {
// Move farest used element out
Node last = tail;
map.remove(last.getKey());
// Remove from list
if (map.size() == 1) {
head = current;
tail = current;
} else {
Node oldHead = head;
current.next = oldHead;
oldHead.prev = current;
head = current;
tail = tail.prev;
tail.next = null;
} } else {
// Add new node to list
if (map.size() == 1) {
head = current;
tail = current;
} else {
Node oldHead = head;
current.next = oldHead;
oldHead.prev = current;
head = current;
}
}
}
}
}

LRU Cache 解答的更多相关文章

  1. [LeetCode]LRU Cache有个问题,求大神解答【已解决】

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

  2. [LeetCode] LRU Cache 最近最少使用页面置换缓存器

    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the fol ...

  3. 【leetcode】LRU Cache

    题目简述: Design and implement a data structure for Least Recently Used (LRU) cache. It should support t ...

  4. LeetCode:LRU Cache

    题目大意:设计一个用于LRU cache算法的数据结构. 题目链接.关于LRU的基本知识可参考here 分析:为了保持cache的性能,使查找,插入,删除都有较高的性能,我们使用双向链表(std::l ...

  5. LRU Cache实现

    最近在看Leveldb源码,里面用到LRU(Least Recently Used)缓存,所以自己动手来实现一下.LRU Cache通常实现方式为Hash Map + Double Linked Li ...

  6. 【leetcode】LRU Cache(hard)★

    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the fol ...

  7. [LintCode] LRU Cache 缓存器

    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the fol ...

  8. LRU Cache [LeetCode]

    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the fol ...

  9. 43. Merge Sorted Array && LRU Cache

    Merge Sorted Array OJ: https://oj.leetcode.com/problems/merge-sorted-array/ Given two sorted integer ...

随机推荐

  1. Razor Generator 将cshtml自动生成对应的CS文件

  2. Spring HTTP invoker 入门

    一.简介 Spring开发团队意识到RMI服务和基于HTTP的服务(如,Hessian)之间的空白.一方面,RMI使用JAVA标准的对象序列化机制,很难穿透防火墙.另一方面,Hessian/Burla ...

  3. Git源码管控规范

    Git分支示意圖 Master:主分支.形成稳定的版本时,才将代码合并到Master分支 Relase:网站发布的分支.通过验证的Bug和功能需求,才合并到Release分支,并将稳定的版本进行备份 ...

  4. 文件系统 busybox and initramfs

    1.busybox制作根文件系统 http://wenku.baidu.com/link?url=h2m_xrj6OsLiHVVhMY2e0C7WKikw_H3dZY_b4mUiW1E7AEf_q34 ...

  5. [转]Laravel 4之请求

    Laravel 4之请求 http://dingjiannan.com/2013/laravel-request/ 获取请求数据 获取当前请求所包括的所有GET和POST数据 Route::get(' ...

  6. 总结oninput、onchange与onpropertychange事件的用法和区别 书写搜索的神奇代码

    总结oninput.onchange与onpropertychange事件的用法和区别 最近手机开发一个模糊搜索的功能组建,在网上就找到这篇文章! 前端页面开发的很多情况下都需要实时监听文本框输入,比 ...

  7. jQuery validate 的valid()方法一直返回true

    1 调用$('#myForm').valid(),一直返回ture eg:html <form id="myForm"> <input class="f ...

  8. 基础命名空间:序列化 System.Runtime.Serialization

    对象通常都有状态(state),从一个对象中抽取这种状态,不论是将它存储于某地,还是通过网络传送,这种抽取动作称为“将一个对象序列化”,而反向处理过程,从一个被序列化的状态重建一个对象即为反序列化. ...

  9. iOS-设计模式之代理反向传值

    代理设计模式就是自己的方法自己不实现,让代理对象去实现. 可以让多个类实现一组方法. 委托模式的好处在于: 1.避免子类化带来的过多的子类以及子类与父类的耦合 2.通过委托传递消息机制实现分层解耦 代 ...

  10. CGFloat和float

    CGFloat :在mac上自适应,在64位的系统,会变宽,32位会变窄,手机没变化float:没有变化