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

  1. [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 ...

  2. [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 ...

  3. [LeetCode#253] Meeting Rooms II

    Problem: Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2] ...

  4. [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 ...

  5. 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],.. ...

  6. [LC] 253. Meeting Rooms II

    Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si ...

  7. 【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 ...

  8. [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 ...

  9. 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

  1. Delphi XE5教程10:Delphi字符集

    内容源自Delphi XE5 UPDATE 2官方帮助<Delphi Reference>,本人水平有限,欢迎各位高人修正相关错误!也欢迎各位加入到Delphi学习资料汉化中来,有兴趣者可 ...

  2. 编写php拓展实例--slime项目(用户登录会话类)

      最近公司换了yaf框架,突然对用c实现php拓展感兴趣了,如果一个功能已经很稳定很成熟而且用的地方很多,那么我们就可以尝试用拓展实现(不一定每种情况都可以写成拓展),写成拓展后就不用每次用都包含一 ...

  3. String对象中常用的方法

    String对象中常用的方法   1.charCodeAt方法返回一个整数,代表指定位置字符的Unicode编码.strObj.charCodeAt(index)说明:index将被处理字符的从零开始 ...

  4. 路由器开发板上的TTL线连接方法

    手头有个MTK双频路由器的开发板,做工良好,但让人蛋疼的是,TTL线没有标注TX/RX/GND/VCC,这个小细节的缺失给使用带来了巨大麻烦.   网上搜了半天也没找到相关电路图,只好遍历测试找到正确 ...

  5. 移植linux4.7.2与ubifs到jz2440

    前言 整个暑假跟着韦东山的视频和书籍移植了linux2.3.6到jz2440,现在自己尝试移植linux4.7.2到板子上,并使用ubifs文件系统代替旧的jffs2文件系统. 下载交叉编译工具链 工 ...

  6. Flash设置全屏后,放到网页中显示不正常

    stage.displayState = StageDisplayState.FULL_SCREEN;//全屏,注意当设置全屏后,放到网页中显示不正常

  7. SSAO

    http://blog.csdn.net/xoyojank/article/details/5734537 http://john-chapman-graphics.blogspot.com/2013 ...

  8. SCI杂志更名时,如何计算影响因子?

  9. 01-04-01【Nhibernate (版本3.3.1.4000) 出入江湖】原生的SQL查询

    Nhibernate 支持原生的SQL查询 : /// <summary> /// 使用原生的SQL查询 /// </summary> /// <param name=& ...

  10. netbean使用技巧

    1.让代码智能提示 有些情况下Ctrl+Space这个键被一些输入法占了,我们需要修改一下点击 工具->常规->快捷键映射->找到显示代码完成弹出式菜单->编辑为你喜欢的键就好 ...