meeting room I & II

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

For example, Given [[0, 30],[5, 10],[15, 20]], return false.

这题和求解有多少架飞机在空中一样。

 public class Solution {
     public boolean canAttendMeetings(Interval[] intervals) {
         List<TimePoint> list = new ArrayList<TimePoint>();

         for (Interval interval : intervals) {
            list.add(new TimePoint(interval.start, true));
            list.add(new TimePoint(interval.end, false));
         }

         Collections.sort(list, new Comparator<TimePoint>() {
             public int compare(TimePoint t1, TimePoint t2) {
                 if (t1.time < t2.time) {
                     return -;
                 } else if (t1.time > t2.time) {
                     return ;
                 } else {
                     if (t1.isStart) {
                         return ;
                     } else {
                         return -;
                     }
                 }
             }
         });
         int count = ;
         for (TimePoint t : list) {
             if (t.isStart) {
                 count++;
                 if (count == ) return false;
             } else {
                 count--;
             }
         }
         return true;
     }
 }

 class TimePoint {
     int time;
     boolean isStart;

     public TimePoint(int time, boolean isStart) {
         this.time = time;
         this.isStart = isStart;
     }
 }

网上看到一个更简单的方法：先sort intervals, 然后看俩个相邻的点是否有重合。

 public boolean canAttendMeetings(Interval[] intervals) {
     Arrays.sort(intervals, new Comparator<Interval>(){
         public int compare(Interval a, Interval b){
             return a.start-b.start;
         }
     });

     for(int i=; i<intervals.length-; i++){
         if(intervals[i].end>intervals[i+].start){
             return false;
         }
     }
     return true;
 }

Meeting Rooms II

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.

方法和第一种相似。

 public class Solution {
     public int minMeetingRooms(Interval[] intervals) {
         List<TimePoint> list = new ArrayList<TimePoint>();

         for (Interval interval : intervals) {
             list.add(new TimePoint(interval.start, true));
             list.add(new TimePoint(interval.end, false));
         }

         Collections.sort(list, (t1, t2) -> {
             if (t1.time < t2.time) {
                 return -;
             } else if (t1.time > t2.time) {
                 return ;
             } else {
                 if (t1.isStart) {
                     return ;
                 } else {
                     return -;
                 }
             }
         });

         int count = , max = ;
         for (TimePoint t : list) {
             if (t.isStart) {
                 count++;
                 max = Math.max(count, max);
             } else {
                 count--;
             }
         }
         return max;
     }
 }

 class TimePoint {
     int time;
     boolean isStart;

     public TimePoint(int time, boolean isStart) {
         this.time = time;
         this.isStart = isStart;
     }
 }

follow up: group all meetings by meeting rooms. 我的解法是对所有会议急于start排序，然后按顺序的分组搞定。