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.

The cache is initialized with a positive capacity.

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
class LRUCache {
private int capacity;
private Map<Integer, Node> map;
private Node head;
private Node tail;
// hash map makes O(1) to get and put
// doubly LL makes removing easier
public LRUCache(int capacity) {
this.map = new HashMap<>();
this.capacity = capacity;
this.head = null;
this.tail = null;
} public int get(int key) {
Node curNode = map.get(key);
if (curNode == null) {
return -1;
} if (tail != curNode) {
if (curNode == head) {
head = head.next;
} else {
curNode.prev.next = curNode.next;
curNode.next.prev = curNode.prev;
}
tail.next = curNode;
curNode.prev = tail;
tail = curNode;
}
return curNode.value;
} public void put(int key, int value) {
Node curNode = map.get(key);
if (curNode != null) {
       // update the current Node, for get(), similiar with this snippet
curNode.value = value;
if (tail != curNode) {
if (curNode == head) {
head = head.next;
} else {
curNode.prev.next = curNode.next;
curNode.next.prev = curNode.prev;
}
tail.next = curNode;
curNode.prev = tail;
tail = curNode;
}
} else {
Node newNode = new Node(key, value);
if (capacity == 0) {
Node tmp = head;
head = tmp.next;
map.remove(tmp.key);
capacity += 1;
}
if (head == null && tail == null) {
head = newNode;
} else {
tail.next = newNode;
newNode.prev = tail;
}
tail = newNode;
capacity -= 1;
map.put(key, newNode);
}
}
} class Node {
int key;
int value;
Node prev;
Node next; public Node(int key, int value) {
this.key = key;
this.value = value;
}
} /**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/

[LC] 146. LRU Cache的更多相关文章

  1. leetcode 146. LRU Cache 、460. LFU Cache

    LRU算法是首先淘汰最长时间未被使用的页面,而LFU是先淘汰一定时间内被访问次数最少的页面,如果存在使用频度相同的多个项目,则移除最近最少使用(Least Recently Used)的项目. LFU ...

  2. [LeetCode] 146. 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. Java for LeetCode 146 LRU Cache 【HARD】

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

  4. leetcode 146. LRU Cache ----- java

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

  5. leetcode@ [146] LRU Cache (TreeMap)

    https://leetcode.com/problems/lru-cache/ Design and implement a data structure for Least Recently Us ...

  6. 146. LRU Cache

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

  7. 【LeetCode】146. LRU Cache

    LRU Cache Design and implement a data structure for Least Recently Used (LRU) cache. It should suppo ...

  8. 146. LRU Cache (List, HashTable)

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

  9. [LeetCode] 146. LRU Cache 近期最少使用缓存

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

随机推荐

  1. Unity 可重复随机数

    出处 https://blogs.unity3d.com/cn/2015/01/07/a-primer-on-repeatable-random-numbers/   (英文原版) http://ww ...

  2. React之Redux知识点补充

    一. reducer纯函数方便自动化测试 二.

  3. Python Email发送,通知业务完成

    Email 发送 #!/usr/bin/python # -*- coding: UTF-8 -*- import base64 import smtplib from email.mime.text ...

  4. POJ 2362:Square 觉得这才算深度搜索

    Square Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 21821   Accepted: 7624 Descripti ...

  5. PHP集成环境wamp和navicat for mysql 的安装

    1. PHP集成环境WAMP的安装: 下载wamp: 链接:https://pan.baidu.com/s/1zvoPEbLdG7y04WWHNM6UcA  密码:mxd3 安装过程如下图: 安装完成 ...

  6. QMainWindow的空间布局结构

    简单讲一下Qt的QMainWindow的结构,Qt的顶级窗口有三种类型,首先是万恶之源(...应该说是大部分控件的父类...)的QWidget,然后是QMainWidget和QDialog,后面两者也 ...

  7. git 一些操作

    1. 代码相关 克隆代码 git clone xxx.git 拉取代码 git pull 查看 修改的 状态 git status 推送代码 git push add 或者 修改代码之后 回滚到 未修 ...

  8. C++queue的使用

    C++队列是一种容器适配器,提供了一种先进先出的数据结构. 队列(queue)模板类定义在<queue>头文件中 基本操作: 定义一个queue变量:queue<Type> q ...

  9. runlevel 运行级别

    linux启动过程 关于Ubuntu 12.04修改默认运行级别,启动字符界面的个人理解   网上通常的做法是:(亲自试验,不管用),如果想直接操作请看绿色字体部分 (1)第一种方法:   由于Red ...

  10. CMake命令之export

    CMake中与export()相关的命令 (注:红色字体是标题,粉色是需要特别需要注意的地方) 总的来说,export()命令想要做的事情可以用一句话概括:Export targets from th ...