253 Meeting Rooms II

253. Meeting Rooms II

1. Question

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]], return2.

2. Implementation

(1) Heap

思路: 这题的本质是问重复的区间有多少个，建立最小堆存放每个interval的结束时间，扫一遍intervals数组，如果最小堆的堆顶代表的时间点小于或等于当前intervals的开始时间，说明有区间不和当前开始时间代表的区间重复，则最小堆删除堆顶。最后看堆的大小判断最少需要多少个meeting rooms

/**
 * 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; }
 * }
 */
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<Integer> endTime = new PriorityQueue<>();
        endTime.add(intervals[0].end);

        for (int i = 1; i < intervals.length; i++) {
            if (endTime.size() > 0 && endTime.peek() <= intervals[i].start) {
                endTime.remove();
            }
            endTime.add(intervals[i].end);
        }
        return endTime.size();
    }
}

3. Time & Space Complexity

Heap: 时间复杂度: O(nlogn)， 空间复杂度O(n)