253. Meeting Rooms II
题目:
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...]
(si < ei), find the minimum number of conference rooms required.
For example,
Given [[0, 30],[5, 10],[15, 20]]
,
return 2
.
链接: http://leetcode.com/problems/meeting-rooms-ii/
题解:
给定一个interval数组,求最少需要多少间教室。初始想法是扫描线算法sweeping-line algorithm,先把数组排序,之后维护一个min-oriented heap。遍历排序后的数组,每次把interval[i].end加入到heap中,然后比较interval.start与pq.peek(),假如interval[i].start >= pq.peek(),说明pq.peek()所代表的这个meeting已经结束,我们可以从heap中把这个meeting的end time移除,继续比较下一个pq.peek()。比较完毕之后我们尝试更新maxOverlappingMeetings。 像扫描线算法和heap还需要好好复习, 直线,矩阵的相交也可以用扫描线算法。
Time Complexity - O(nlogn), Space Complexity - O(n)
/**
* 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 Solution {
public int minMeetingRooms(Interval[] intervals) {
if(intervals == null || intervals.length == 0)
return 0; Arrays.sort(intervals, new Comparator<Interval>() {
public int compare(Interval t1, Interval t2) {
if(t1.start != t2.start)
return t1.start - t2.start;
else
return t1.end - t2.end;
}
}); int maxOverlappingMeetings = 0;
PriorityQueue<Integer> pq = new PriorityQueue<>(); // min oriented priority queue for(int i = 0; i < intervals.length; i++) { // sweeping-line algorithms
pq.add(intervals[i].end);
while(pq.size() > 0 && intervals[i].start >= pq.peek())
pq.remove(); maxOverlappingMeetings = Math.max(maxOverlappingMeetings, pq.size());
} return maxOverlappingMeetings;
}
}
二刷:
二刷参考了@pinkfloyda的写法。对start以及end排序,然后再两个数组中对end和start进行比较。代码很简洁,速度也很快,非常值得学习。
Java:
Time Complexity - O(nlogn), Space Complexity - O(n)
/**
* 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 Solution {
public int minMeetingRooms(Interval[] intervals) {
if (intervals == null || intervals.length == 0) return 0;
int len = intervals.length;
int[] starts = new int[len];
int[] ends = new int[len];
for (int i = 0; i < len; i++) {
starts[i] = intervals[i].start;
ends[i] = intervals[i].end;
}
Arrays.sort(starts);
Arrays.sort(ends); int minRooms = 0, endIdx = 0;
for (int i = 0; i < len; i++) {
if (starts[i] < ends[endIdx]) minRooms++;
else endIdx++;
} return minRooms;
}
}
三刷:
class Solution {
public int minMeetingRooms(int[][] intervals) {
if (intervals == null || intervals.length == 0) return 0;
Arrays.sort(intervals, (int[] i1, int[] i2) ->
i1[0] != i2[0] ? i1[0] - i2[0] : i1[1] - i2[1]);
Queue<Integer> q = new PriorityQueue<>();
int result = 0;
for (int[] interval : intervals) {
while (!q.isEmpty() && q.peek() <= interval[0]) q.poll();
q.offer(interval[1]);
result = Math.max(result, q.size());
}
return result;
}
}
Reference:
https://leetcode.com/discuss/71846/super-easy-java-solution-beats-98-8%25
https://leetcode.com/discuss/50911/ac-java-solution-using-min-heap
https://leetcode.com/discuss/82292/explanation-super-easy-java-solution-beats-from-%40pinkfloyda
https://leetcode.com/discuss/70998/java-ac-solution-greedy-beats-92-03%25
253. Meeting Rooms II的更多相关文章
- [LeetCode] 253. Meeting Rooms II 会议室之二
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...
- [LeetCode] 253. Meeting Rooms II 会议室 II
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...
- [LeetCode#253] Meeting Rooms II
Problem: Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2] ...
- [leetcode]253. Meeting Rooms II 会议室II
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...
- 253. Meeting Rooms II 需要多少间会议室
[抄题]: Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],.. ...
- [LC] 253. Meeting Rooms II
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...
- 【LeetCode】253. Meeting Rooms II 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 排序+堆 日期 题目地址:https://leetco ...
- [LeetCode] Meeting Rooms II 会议室之二
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...
- LeetCode Meeting Rooms II
原题链接在这里:https://leetcode.com/problems/meeting-rooms-ii/ Given an array of meeting time intervals con ...
随机推荐
- Translation002—Package Index(Android包索引)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 看本翻译前请您注意: 本人初学android,可能有的翻译不是非常准确,但本人尽最大努力,不清楚处会做标记,并附 ...
- Microsoft server software support for Microsoft Azure virtual machines
http://support.microsoft.com/kb/2721672/en-us Article ID: 2721672 - Last Review: November 22, 2014 ...
- OpenNMS架构介绍
一.OpenNMS简介 OpenNMS的开发基于TMN及FCAPS这两个模型. 电信管理网络(TMN)是由 ITU-T 推荐 M.3000于1985年提出作为一种应用于电信服务供应商所持有的运营支持系 ...
- Ubuntu中设置环境变量详解
1, 为单一用户:.bashrc: 为每一个运行bash shell的用户执行此文件.当bash shell被打开时,该文件被读取.打开用户主目录下的.bashrc,在这个文件中加入export PA ...
- 与MySQL交互(felixge/node-mysql)
目录 简介和安装 测试MySQL 认识一下Connection Options MYSQL CURD 插入 更新 查询 删除 Nodejs 调用带out参数的存储过程,并得到out参数返回值 结束数据 ...
- Unity3D实现立体迷宫寻宝
Unity3D实现立体迷宫寻宝 这个小游戏是一个白痴在一个昏暗的房间走动找到关键得分点,然后通关游戏.入门Unity3D做的第一款游戏,比较无聊,但实现了一般的游戏功能.如,人物控制,碰撞检测,主控制 ...
- SQL_SERVER_2008升级SQL_SERVER_2008_R2办法 (一、升级;二、重新xie载安装)
SQL_SERVER_2008升级SQL_SERVER_2008_R2两种办法 今天将由于需要就将我的SQL 2008升级到SQL 2008 R2. 说到为什么要升级是因为,因附加数据库时发现报错 ...
- 多线程 1-pthread 和NSThread
一.基本内容介绍: 进程: 正在运行的程序就叫进程 每个进程之间是相互独立的,每个进程均运行在其专用且受保护的内存空间内. 线程: 在程序内工作的基本执行单元(每个进程至 ...
- 【个人笔记】003-PHP基础-01-PHP快速入门-03-PHP环境搭建
003-PHP基础-01-PHP快速入门 03-PHP环境搭建 1.客户端(浏览器) IE FireFox CHROME Opera Safari 2.服务器 是运行网站的基本 是放置程序代码的地方 ...
- android studio 突然出现Gradle project sync failed 错误
出现: 之前还是好好的,突然就出现Gradle project sync failed 错误,网上原因可能是工具的问题. 解决办法: 重新打开android studio就好了.不知道大家还有其他的 ...