原题链接在这里:https://leetcode.com/problems/design-log-storage-system/description/

题目:

You are given several logs that each log contains a unique id and timestamp. Timestamp is a string that has the following format: Year:Month:Day:Hour:Minute:Second, for example, 2017:01:01:23:59:59. All domains are zero-padded decimal numbers.

Design a log storage system to implement the following functions:

void Put(int id, string timestamp): Given a log's unique id and timestamp, store the log in your storage system.

int[] Retrieve(String start, String end, String granularity): Return the id of logs whose timestamps are within the range from start to end. Start and end all have the same format as timestamp. However, granularity means the time level for consideration. For example, start = "2017:01:01:23:59:59", end = "2017:01:02:23:59:59", granularity = "Day", it means that we need to find the logs within the range from Jan. 1st 2017 to Jan. 2nd 2017.

Example 1:

put(1, "2017:01:01:23:59:59");
put(2, "2017:01:01:22:59:59");
put(3, "2016:01:01:00:00:00");
retrieve("2016:01:01:01:01:01","2017:01:01:23:00:00","Year"); // return [1,2,3], because you need to return all logs within 2016 and 2017.
retrieve("2016:01:01:01:01:01","2017:01:01:23:00:00","Hour"); // return [1,2], because you need to return all logs start from 2016:01:01:01 to 2017:01:01:23, where log 3 is left outside the range.

Note:

  1. There will be at most 300 operations of Put or Retrieve.
  2. Year ranges from [2000,2017]. Hour ranges from [00,23].
  3. Output for Retrieve has no order required.

题解:

Inorder to retireve range of timestamp. It needs a linear data structure to store log based on timestamp. It could be list or array, but every time, it needs to retrieve everything again.

Or we could use TreeMap. Thus timestamp is sorted.

When we have the start and end time, revise a little bit from the granularity index.

Use TreeMap subMap to get sub map range from [start, end] timestamp.

And add values to res.

Time Complexity: put, O(logn). n = current count of log. retrieve(logn + m). m = count of logs between start and end.

Space: O(n).

AC Java:

 class LogSystem {
private TreeMap<String, HashSet<Integer>> treeMap;
private HashMap<String, Integer> map;
private String min = "2000:01:01:00:00:00";
private String max = "2017:12:31:23:59:59"; public LogSystem() {
this.treeMap = new TreeMap<>();
this.map = new HashMap<>();
map.put("Year", 4);
map.put("Month", 7);
map.put("Day", 10);
map.put("Hour", 13);
map.put("Minute", 16);
map.put("Second", 19);
} public void put(int id, String timestamp) {
treeMap.putIfAbsent(timestamp, new HashSet<>());
treeMap.get(timestamp).add(id);
} public List<Integer> retrieve(String s, String e, String gra) {
int index = map.get(gra);
String start = s.substring(0, index) + min.substring(index);
String end = e.substring(0, index) + max.substring(index); Map<String, HashSet<Integer>> subMap = treeMap.subMap(start, true, end, true);
List<Integer> res = new ArrayList<>();
for(Map.Entry<String, HashSet<Integer>> entry : subMap.entrySet()){
res.addAll(entry.getValue());
} return res;
}
} /**
* Your LogSystem object will be instantiated and called as such:
* LogSystem obj = new LogSystem();
* obj.put(id,timestamp);
* List<Integer> param_2 = obj.retrieve(s,e,gra);
*/

用list直接存log. 根据granularity取出timestamp的substring直接比较.

Time Complexity: put, O(1). retrieve, O(logs.size()).

Space: O(logs.size()).

AC Java:

 class LogSystem {
List<String []> logs;
List<String> units = Arrays.asList("Year", "Month", "Day", "Hour", "Minute", "Second");
int [] indices = new int[]{4,7,10,13,16,19}; public LogSystem() {
logs = new LinkedList<String []>();
} public void put(int id, String timestamp) {
logs.add(new String[]{Integer.toString(id), timestamp});
} public List<Integer> retrieve(String s, String e, String gra) {
List<Integer> res = new ArrayList<Integer>();
int ind = indices[units.indexOf(gra)]; String sSub = s.substring(0, ind);
String eSub = e.substring(0, ind); for(String [] log : logs){
String sub = log[1].substring(0, ind);
if(sub.compareTo(sSub)>=0 && sub.compareTo(eSub)<=0){
res.add(Integer.valueOf(log[0]));
}
}
return res;
}
} /**
* Your LogSystem object will be instantiated and called as such:
* LogSystem obj = new LogSystem();
* obj.put(id,timestamp);
* List<Integer> param_2 = obj.retrieve(s,e,gra);
*/

LeetCode Design Log Storage System的更多相关文章

  1. [LeetCode] Design Log Storage System 设计日志存储系统

    You are given several logs that each log contains a unique id and timestamp. Timestamp is a string t ...

  2. Design Log Storage System

    You are given several logs that each log contains a unique id and timestamp. Timestamp is a string t ...

  3. [leetcode-635-Design Log Storage System]

    You are given several logs that each log contains a unique id and timestamp. Timestamp is a string t ...

  4. [LeetCode] Design Search Autocomplete System 设计搜索自动补全系统

    Design a search autocomplete system for a search engine. Users may input a sentence (at least one wo ...

  5. [LeetCode] Design In-Memory File System 设计内存文件系统

    Design an in-memory file system to simulate the following functions: ls: Given a path in string form ...

  6. Bigtable: A Distributed Storage System for Structured Data

    https://static.googleusercontent.com/media/research.google.com/en//archive/bigtable-osdi06.pdf Abstr ...

  7. Storage System and File System Courses

    I researched a lot about storage system classes given at good universities this year. This had two r ...

  8. 1.1 Introduction中 Kafka as a Storage System官网剖析(博主推荐)

    不多说,直接上干货! 一切来源于官网 http://kafka.apache.org/documentation/ Kafka as a Storage System kafka作为一个存储系统 An ...

  9. Blockstack: A Global Naming and Storage System Secured by Blockchains

    作者:Muneeb Ali, Jude Nelson, Ryan Shea, and Michael Freedman Blockstack Labs and Princeton University ...

随机推荐

  1. html4与html5的区别

    一.HTML5更加灵活,支持下列多种形式 1.标签名可以大写(不推荐) -<SpAN>这个HTML5也的认</SpAN> 2.属性双引号可选(推荐添加双引号) -<div ...

  2. 求最长不下降子序列(nlogn)

    最长递增子序列问题:在一列数中寻找一些数,这些数满足:任意两个数a[i]和a[j],若i<j,必有a[i]<a[j],这样最长的子序列称为最长递增子序列. 设dp[i]表示以i为结尾的最长 ...

  3. idea软件编码已经设置好了为utf-8,但是svn中down下来的文件格式本身不是utf-8的,此时打开后会出现中文乱码解决方法

    我是个idea的忠实用户,新公司的项目都是用eclipse做的,通过svn拉下代码后发现,注释的内容里,中文内容都是乱码.问过项目负责人,说可能是GBK编码. 但是,我通过idea的setting设置 ...

  4. TNS-12541: TNS:no listener , TNS-12542: TNS:address already in use

    查看数据库监听状态不对$ lsnrctl status LSNRCTL for IBM/AIX RISC System/6000: Version 10.2.0.5.0 - Production on ...

  5. hibernate 实体对象的三种状态以及转换关系。

    最新的Hibernate文档中为Hibernate对象定义了四种状态(原来是三种状态,面试的时候基本上问的也是三种状态),分别是:瞬时态(new, or transient).持久态(managed, ...

  6. 使用VirtualBox安装Android 4.2.2 x86 .

    http://blog.csdn.net/kunoy/article/details/8768205 virtual box 安装 android x86 不显示鼠标, --> 控制 --> ...

  7. NumPy切片和索引

    NumPy - 切片和索引 ndarray对象的内容可以通过索引或切片来访问和修改,就像 Python 的内置容器对象一样. 如前所述,ndarray对象中的元素遵循基于零的索引. 有三种可用的索引方 ...

  8. Django进阶Template篇001 - 常用模板标签及过滤器

    一.模板的组成 HTML代码+逻辑控制代码 二.逻辑控制代码的组成 1.变量(使用双大括号来引用变量) {{ var_name }} 2.标签(tag)的使用(使用大括号和百分号的组成来表示使用tag ...

  9. 使用Selenium通过浏览器对网站进行自动化测试和相关问题

    使用Selenium通过浏览器对网站进行自动化测试 自动化测试概念: 一般是指软件测试的自动化,软件测试就是在预设条件下运行系统或应用程序,评估运行结果,预先条件应包括正常条件和异常条件. 广义上来讲 ...

  10. CPU Usage (C#) 测试

    注意:算法仅供参考. cpuusage.cs using System; using System.Collections.Generic; using System.Diagnostics; usi ...