[LeetCode] 253. Meeting Rooms II

mengera88 发布于2019-08-19 11:14 / 1613人阅读

Problem

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.

Example 1:

Input: [[0, 30],[5, 10],[15, 20]]
Output: 2
Example 2:

Input: [[7,10],[2,4]]
Output: 1

Solution Using Array

</>复制代码 
class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        //0, 100 ---- 1
        //0, 200 ---- 2
        //0, 300 ---- 3
        //150, 250 -- 3
        //160, 260 -- 4
        //210, 310 -- 4
        int n = intervals.length;
        int[] startTimes = new int[n];
        int[] endTimes = new int[n];
        for (int i = 0; i < n; i++) {
            startTimes[i] = intervals[i].start;
            endTimes[i] = intervals[i].end;
        }
        Arrays.sort(startTimes);
        Arrays.sort(endTimes);
        int count = 0;
        int i = 0, j = 0;
        while (i < n && j < n) {
            //once overlaps, room++
            if (startTimes[i] < endTimes[j]) count++;
            //not overlapping, release prev meeting room
            else j++;
            i++;
        }
        return count;
    }
}

Using Heap

</>复制代码 
class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) return 0;
        Arrays.sort(intervals, (a, b) -> a.start-b.start);
        PriorityQueue queue = new PriorityQueue<>((a, b) -> a.end-b.end);
        /*add room for the first meeting*/
        queue.add(intervals[0]);
        for (int i = 1; i < intervals.length; i++) {
            Interval pre = queue.poll();
            Interval cur = intervals[i];
            
            // if new meeting time overlapped, add a new room for this meeting
            if (cur.start < pre.end) queue.offer(cur);
            
            // if not overlapped, no new room for the new meeting, 
            // just update the ending time for the earliest-to-end room
            else pre.end = cur.end;
            
            // put this previous earliest-to-end room back back to heap, 
            // so that in next iteration, we get the new earliest-to-end room
            queue.offer(pre);
        }
        return queue.size();
    }
}

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