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.

思路：标准扫描线算法，start event + 1, end event -1, 把所有的event点按照时间sort一下，如果时间相同，-1排在1前面，O(NlogN) + O(N).

class Solution {
    private class Node {
        public int time;
        public int flag;
        public Node(int time, int flag) {
            this.time = time;
            this.flag = flag;
        }
    }
    
    public int minMeetingRooms(int[][] intervals) {
        if(intervals == null || intervals.length == 0) {
            return 0;
        }    
        List<Node> list = new ArrayList<>();
        for(int[] interval: intervals) {
            list.add(new Node(interval[0], 1));
            list.add(new Node(interval[1], -1));
        }
        
        int count = 0;
        int maxcount = 0;
        Collections.sort(list, (a, b) -> (a.time != b.time ? a.time - b.time : a.flag - b.flag));
        for(Node node: list) {
            if(node.flag == 1) {
                count++;
            } else {
                count--;
            }
            maxcount = Math.max(maxcount, count);
        }
        return maxcount;
    }
}

也可以用treemap写，<Integer, Integer> 分别代表的是start 和出现的次数. start的时候，+1, end的时候，-1；然后过keyset，就是扫描一遍，求最大；

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        TreeMap<Integer, Integer> treemap = new TreeMap<>();
        for(int[] interval: intervals) {
            treemap.put(interval[0], treemap.getOrDefault(interval[0], 0) + 1);
            treemap.put(interval[1], treemap.getOrDefault(interval[1], 0) - 1);
        }
        
        int maxcount = 0;
        int count = 0;
        for(Integer key: treemap.keySet()) {
            count += treemap.get(key);
            maxcount = Math.max(maxcount, count);
        }
        return maxcount;
    }
}