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. c++ 13

    一.向量 ... 10.size/resize/clear/capacity/reserve 1)向量的大小可增可减,使向量大小改变的函数包括:resize/push_back/pop_back/cl ...

  2. web前端之 JS

    JavaScript概述 JavaScript是一门编程语言,简称js,由浏览器编译并运行,JS说白了就是让页面能够动起来 js存在形式 1.在html页面中 <script> alert ...

  3. web前端之 HTML介绍

    概述 HTML是英文Hyper Text Mark-up Language(超文本标记语言)的缩写,他是一种制作万维网页面标准语言(标记).相当于定义统一的一套规则,大家都来遵守他,这样就可以让浏览器 ...

  4. 0..n去掉一个数,给你剩下的数,找出去掉的那个数

    转载请注明转自blog.csdn.net/souldak , 微博@evagle 首先,考虑没有去掉那些数,如果n是奇数,n+1个最低位肯定是0101...01,count(0)=count(1),如 ...

  5. .NET批量大数据插入性能分析及比较

    数据插入使用了以下几种方式 1. 逐条数据插入2. 拼接sql语句批量插入3. 拼接sql语句并使用Transaction4. 拼接sql语句并使用SqlTransaction5. 使用DataAda ...

  6. sqlite3安装

    SQLite命令行程序(CLP)是开始使用SQLite的最好选择,按照如下步骤获取CLP: 1).打开浏览器进入SQLite主页,www.sqlite.org. 2).单击页面顶部的下载链接(Down ...

  7. js将对象转成字符串-支持微信

    最近写一个微信项目时用到了 把对象转成字符串,因为我需要把它存在cookie中,碰到了一些问题,在这里分享一下. 要转换的就是这货~ var FBinf = { "workPlacesCod ...

  8. FlexSlider是一个非常出色的jQuery滑动切换插件

    FlexSlider是一个非常出色的jQuery滑动切换插件,它支持所有主流浏览器,并有淡入淡出效果.适合所有初级和高级网页设计师使用.不过很多人都只是使用默认的参数,今天来说说具体的参数来给大家看看 ...

  9. html进阶css(3)

    css的某些样式是具有继承性的,那么什么是继承呢?继承是一种规则,它允许格式不仅应用于某个特定html标签元素,而且应用于其后代. <!doctype html> <html> ...

  10. Android应用去掉标题栏的方法

    1.在代码里实现 this.requestWindowFeature(Window.FEATURE_NO_TITLE);//去掉标题栏,this指当前的Activity 这句代码一定要加在setCon ...