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253. 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.
//https://discuss.leetcode.com/topic/35253/explanation-of-super-easy-java-solution-beats-98-8-from-pinkfloyda
public class Solution {
public int minMeetingRooms(Interval[] intervals) {
int[] starts = new int[intervals.length];
int[] ends = new int[intervals.length];
for(int i=0; i<intervals.length; i++) {
starts[i] = intervals[i].start;
ends[i] = intervals[i].end;
}
Arrays.sort(starts);
Arrays.sort(ends);
int rooms = 0;
int endsItr = 0;
for(int i=0; i<starts.length; i++) {
if(starts[i]<ends[endsItr])
rooms++;
else
endsItr++;
}
return rooms;
}
}
//////////////////////////////////////////
//https://discuss.leetcode.com/topic/20958/ac-java-solution-using-min-heap
public int minMeetingRooms(Interval[] intervals) {
if (intervals == null || intervals.length == 0)
return 0;
// Sort the intervals by start time
Arrays.sort(intervals, new Comparator<Interval>() {
public int compare(Interval a, Interval b) { return a.start - b.start; }
});
// Use a min heap to track the minimum end time of merged intervals
PriorityQueue<Interval> heap = new PriorityQueue<Interval>(intervals.length, new Comparator<Interval>() {
public int compare(Interval a, Interval b) { return a.end - b.end; }
});
// start with the first meeting, put it to a meeting room
heap.offer(intervals[0]);
for (int i = 1; i < intervals.length; i++) {
// get the meeting room that finishes earliest
Interval interval = heap.poll();
if (intervals[i].start >= interval.end) {
// if the current meeting starts right after
// there's no need for a new room, merge the interval
interval.end = intervals[i].end;
} else {
// otherwise, this meeting needs a new room
heap.offer(intervals[i]);
}
// don't forget to put the meeting room back
heap.offer(interval);
}
return heap.size();
}