LeetCode LFU Cache
原题链接在这里:https://leetcode.com/problems/lfu-cache/
题目:
Design and implement a data structure for Least Frequently Used (LFU) 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 reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LFUCache cache = new LFUCache( 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.get(3); // returns 3.
cache.put(4, 4); // evicts key 1.
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
题解:
去掉least frequently used element, 就需要一个min来maintain到目前最不被利用的元素的利用次数.
用三个map, 一个是维护正常key value pair的HashMap<Integer, Integer> keyVals.
第二个是维护每个key的使用次数.
第三个是维护每个count下对应的key set.
当put第一个元素时, min=1, 对应更新keyVals, keyCounts 和 countKeySets.
get时, key的count要加一, 对应调整keyCounts 和 countKeySets. 若这个key的count恰巧是最少使用次数的最后一个值,那么最少使用次数min++.
在达到capacity后在加新key时利用min来找到least frequently used element, 并对应调整keyVals, keyCounts 和 countKeySets.
Note: corner case capacity <= 0.
countToKeys need LinkedHashSet, because when there is even, evict the oldest one.
Time Complexity: get, O(1). put, O(1).
Space: O(n).
AC Java:
public class LFUCache {
HashMap<Integer, Integer> keyVals;
HashMap<Integer, Integer> keyCounts;
HashMap<Integer, LinkedHashSet<Integer>> countKeySets;
int capacity;
int min; public LFUCache(int capacity) {
this.capacity = capacity;
this.min = -1;
keyVals = new HashMap<Integer, Integer>();
keyCounts = new HashMap<Integer, Integer>();
countKeySets = new HashMap<Integer, LinkedHashSet<Integer>>();
countKeySets.put(1, new LinkedHashSet<Integer>());
} public int get(int key) {
if(!keyVals.containsKey(key)){
return -1;
}
int count = keyCounts.get(key);
keyCounts.put(key, count+1);
countKeySets.get(count).remove(key);
if(count == min && countKeySets.get(count).size() == 0){
min++;
}
if(!countKeySets.containsKey(count+1)){
countKeySets.put(count+1, new LinkedHashSet<Integer>());
}
countKeySets.get(count+1).add(key);
return keyVals.get(key);
} public void put(int key, int value) {
if(capacity <= 0){
return;
} if(keyVals.containsKey(key)){
keyVals.put(key, value);
get(key);
return;
}
if(keyVals.size() >= capacity){
int leastFreq = countKeySets.get(min).iterator().next();
keyVals.remove(leastFreq);
keyCounts.remove(leastFreq);
countKeySets.get(min).remove(leastFreq);
}
keyVals.put(key, value);
keyCounts.put(key, 1);
countKeySets.get(1).add(key);
min = 1;
}
} /**
* Your LFUCache object will be instantiated and called as such:
* LFUCache obj = new LFUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/
类似LRU Cache, All O`one Data Structure.
LeetCode LFU Cache的更多相关文章
- [LeetCode] LFU Cache 最近最不常用页面置换缓存器
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the f ...
- Leetcode: LFU Cache && Summary of various Sets: HashSet, TreeSet, LinkedHashSet
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the f ...
- [LeetCode] 460. LFU Cache 最近最不常用页面置换缓存器
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the f ...
- 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 ...
- [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 ...
- LFU Cache
2018-11-06 20:06:04 LFU(Least Frequently Used)算法根据数据的历史访问频率来淘汰数据,其核心思想是“如果数据过去被访问多次,那么将来被访问的频率也更高”. ...
- Leetcode:LRU Cache,LFU Cache
在Leetcode上遇到了两个有趣的题目,分别是利用LRU和LFU算法实现两个缓存.缓存支持和字典一样的get和put操作,且要求两个操作的时间复杂度均为O(1). 首先说一下如何在O(1)时间复杂度 ...
- leetcode 460. LFU Cache
hash:存储的key.value.freq freq:存储的freq.key,也就是说出现1次的所有key在一起,用list连接 class LFUCache { public: LFUCache( ...
- [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 ...
随机推荐
- Loadrunder脚本篇——web_custom_request做接口测试
一.POST + JSON格式参数 例: web_custom_request("create", "URL=http://xxx.xxx.x.xx:1600/ditui ...
- SqlHelper简单实现(通过Expression和反射)3.实体,数据传输对象(DTO)Helper类设计
EntityHelper的主要功能有: 1.通过反射获取DTO的字段,主要提供给在需要从Entity获取数据后,填充给DTO并返回的作用: 通过反射获取PropertyInfo[]对象,然后取出Nam ...
- matlab fread
Matlab中fread函数用法 “fread”以二进制形式,从文件读出数据. 语法1:[a,count]=fread(fid,size,precision) 语法2:[a,count]=fre ...
- Django-Ajax进阶
一.Ajax上传文件 1.form表单上传文件 文件和其他的数据类型不一样,是一个二进制的形式 Form上传文件的时候切记要加上:enctype="multipart/form-data&q ...
- React Native的生命周期解析
在React Native中使用组件来封装界面模块时,整个界面就是一个大的组件,开发过程就是不断优化和拆分界面组件.构造整个组件树的过程. 上张图涵盖了一个组件从创建.运行到销毁的整个过程.大家可以看 ...
- centOS安装apache服务器
# yum install httpd 启动 systemctl start httpd 重启 systemctl restart httpd 停止 systemctl stop httpd
- keepalived检测脚本及注意事项
keepalived检测脚本的作用及注意事项: 默认每隔3秒钟执行一次检测脚本,检查nginx服务是否启动,如果没启动就把nginx服务启动起来,如果启动不成功,就把keepalived服务down掉 ...
- MySQL实时性能监控工具doDBA tools
doDBA tools是什么? doDBA tools是一个基于控制台的远程监控工具,它不需要在本地/远程系统上安装任何软件,它可以实时收集操作系统.MySQL.InnoDB的实时性能状态数据,并可以 ...
- 为什么原生的servlet是线程不安全的而Struts2是线程安全的?
因为原生的servlet在整个application生命周期中,只在初次访问的时候实例化一次,以后都不会再实例化,只会调用Server方法进行响应,所以如果在servlet类中定义成员变量,那么就会让 ...
- Hadoop的Docker镜像构建
1.Dockerfile ###Dockerfile -- beagin FROM ubuntu:trusty #MAINTAINER The Hue Team "https://githu ...