Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]
Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

TreeMap 解法:

Use TreeMap to easily find the lower and higher keys, the key is the start of the interval.
Merge the lower and higher intervals when necessary. The time complexity for adding is O(logN) since lowerKey(), higherKey(), put() and remove() are all O(logN). It would be O(N) if you use an ArrayList and remove an interval from it.

Summary of TreeMap

The map is sorted according to the natural ordering of its keys, or by a Comparator provided at map creation time, depending on which constructor is used.

This implementation provides guaranteed log(n) time cost for the containsKeygetput and remove operations.

methods include: ceilingKey(), floorKey(), higherKey(), lowerKey(), firstKey(): return the lowest key in the map, lastKey() return the highest key in the map

 /**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class SummaryRanges {
TreeMap<Integer, Interval> tree; /** Initialize your data structure here. */
public SummaryRanges() {
tree = new TreeMap<>();
} public void addNum(int val) {
if (tree.containsKey(val)) return;
Integer l = tree.lowerKey(val);
Integer h = tree.higherKey(val); //case 1: val is the only number between the two intervals
if (l!=null && h!=null && val==tree.get(l).end+1 && val==h-1) {
tree.get(l).end = tree.get(h).end;
tree.remove(h);
} //case 2 & 3: val is in one interval or is the next elem of that interval's last elem
else if (l!=null && val<=tree.get(l).end+1) {
tree.get(l).end = Math.max(tree.get(l).end, val);
} //case 4: val is the first elem of a interval
else if (h!=null && val==h-1) {
tree.put(val, new Interval(val, tree.get(h).end));
tree.remove(h);
} //case 5: val does not adhere to any interval
else {
tree.put(val, new Interval(val, val));
}
} public List<Interval> getIntervals() {
return new ArrayList<>(tree.values());
}
} /**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* List<Interval> param_2 = obj.getIntervals();
*/

TreeSet 解法:

 /**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class SummaryRanges { /** Initialize your data structure here. */
public SummaryRanges() {
itvlSet = new TreeSet<Interval>(new Comparator<Interval>(){
public int compare(Interval v1, Interval v2){
return v1.start-v2.start;
}
}); } public void addNum(int val) {
Interval itvl = new Interval(val,val);
Interval pre = itvlSet.floor(itvl);
Interval after = itvlSet.ceiling(itvl); if ( (pre!=null && pre.end >= val) || (after!=null && after.start <=val)) return; if (pre!=null && pre.end==val-1){
itvlSet.remove(pre);
itvl.start = pre.start;
}
if (after!=null && after.start==val+1){
itvlSet.remove(after);
itvl.end = after.end;
}
itvlSet.add(itvl);
} public List<Interval> getIntervals() {
return new ArrayList<Interval>(itvlSet); } TreeSet<Interval> itvlSet;
} /**
* Your SummaryRanges object will be instantiated and called as such:
* SummaryRanges obj = new SummaryRanges();
* obj.addNum(val);
* List<Interval> param_2 = obj.getIntervals();
*/

Leetcode: Data Stream as Disjoint Intervals && Summary of TreeMap的更多相关文章

  1. [LeetCode] Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  2. 352[LeetCode] Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  3. leetcode@ [352] Data Stream as Disjoint Intervals (Binary Search & TreeSet)

    https://leetcode.com/problems/data-stream-as-disjoint-intervals/ Given a data stream input of non-ne ...

  4. [LeetCode] 352. Data Stream as Disjoint Intervals 分离区间的数据流

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  5. 【leetcode】352. Data Stream as Disjoint Intervals

    问题描述: Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers ...

  6. [Swift]LeetCode352. 将数据流变为多个不相交间隔 | Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  7. 352. Data Stream as Disjoint Intervals (TreeMap, lambda, heapq)

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  8. 352. Data Stream as Disjoint Intervals

    Plz take my miserable life T T. 和57 insert interval一样的,只不过insert好多. 可以直接用57的做法一个一个加,然后如果数据大的话,要用tree ...

  9. [leetcode]352. Data Stream as Disjoint Intervals

    数据流合并成区间,每次新来一个数,表示成一个区间,然后在已经保存的区间中进行二分查找,最后结果有3种,插入头部,尾部,中间,插入头部,不管插入哪里,都判断一下左边和右边是否能和当前的数字接起来,我这样 ...

随机推荐

  1. REST性能测试方案

    1.REST简介 REST(代表性状态传输,Representational State Transfer)是一种Web服务设计模型.REST定义了一组体系架构原则,您可以根据这些原则设计以系统资源为 ...

  2. jdk Tomcat配置

    安装Tomcat需要先安装JDKJDK安装JDK安装会有两次,两次不能再同一目录 你可以新建两个文件夹 JDK JRE 第一次安装在JDK 第二次在JRE在我的电脑右键属性 高级系统设置 环境变量 系 ...

  3. (转)JAVA 调用matlab

    本文仅用于学习. 原文地址链接:http://blog.csdn.net/wannshan/article/details/5907877 前段时间摸索了java调用matlab东西,不说学的有多深, ...

  4. java ReentrantReadWriteLock

    // read and write lock is mutual exclusion lock //Listing 7-3. Using ReadWriteLock to Satisfy a Dict ...

  5. php四个常用类封装 :MySQL类、 分页类、缩略图类、上传类;;分页例子;

    Mysql类 <?php /** * Mysql类 */ class Mysql{ private static $link = null;//数据库连接 /** * 私有的构造方法 */ pr ...

  6. 基于ace后台管理系统模板--CMS(Thinkphp框架)的筹划

    临近春节,准备自己做一个关于宠物的cms网站,特写下此博客提醒自己,尽量争取在过年前做好.废号少说,先梳理下接下来准备使用的工具.. 由于最近在学习thinkphp,所以打算用这个框架来作为主体,可能 ...

  7. 6 个JavaScript日期处理库

    1. Later.js Later.js, a stadalone JavaScript library, offers an advanced usage for triggering recurr ...

  8. java整合spring和hadoop HDFS

    http://blog.csdn.net/kokjuis/article/details/53586406 http://download.csdn.net/detail/kokjuis/970932 ...

  9. 【android学习2】:Eclipse中HttpServlet类找不到

    Eclipse中使用的HttpServlet类之所以识别不到的原因是没有导入Servlet-api.jar包,这个包在所安装在的tomcat的lib文件下,所以只需要导入即可. 在需要导入的工程上右键 ...

  10. 使用多种客户端消费WCF RestFul服务(一)——服务端

    RestFul风格的WCF既然作为跨平台.跨语言.跨技术的一种方式出现,并且在ASP.NET API流行起来之前还是架构的首选技术之一,那么我们就来简要的介绍一下WCF在各个平台客户端的操作. 开发工 ...