252, 253. 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<>();
        
        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 -1;
                } else if (t1.time > t2.time) {
                    return 1;
                } else {
                    if (t1.isStart) {
                        return 1;
                    } else {
                        return -1;
                    }
                }  
            }
        });
        int count = 0;
        for (TimePoint t : list) {
            if (t.isStart) {
                count++;
                if (count == 2) 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 class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals, (a, b) -> a.start - b.start);

        for (int i = 0; i < intervals.length - 1; i++) {
            if (intervals[i].end > intervals[i + 1].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 -1;
            } else if (t1.time > t2.time) {
                return 1;
            } else {
                if (t1.isStart) {
                    return 1;
                } else {
                    return -1;
                }
            }
        });

        int count = 0, max = 0;
        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。