Meeting Rooms

最新推荐文章于 2024-07-31 16:37:35 发布

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本文介绍了一种使用线扫描算法判断是否存在会议时间冲突的方法。通过定义开始和结束时间点，并利用比较器进行排序，确保了算法的有效性和正确性。最终通过计数器检查是否有超过一个会议在同一时间进行的情况。

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线扫描求解

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.

Have you met this question in a real interview?

Yes

Example

Given intervals = [[0,30],[5,10],[15,20]], return false.

/**
 * Definition of Interval:
 * public classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 * }
 */

class Point {
    int time;
    int flag;
    public Point(int time, int flag) {
        this.time = time;
        this.flag = flag;
    }
}
public class Solution {
    /**
     * @param intervals: an array of meeting time intervals
     * @return: if a person could attend all meetings
     */
    public boolean canAttendMeetings(List<Interval> intervals) {
        // Write your code here
        if (intervals == null || intervals.size() == 0) {
            return true;
        }
        Comparator<Point> cmp = new Comparator<Point>() {
            public int compare(Point p1, Point p2) {
                if (p1.time == p2.time) {
                    return p1.flag - p2.flag;
                }
                return p1.time - p2.time;
            }
        };
        List<Point> list = new ArrayList<>();
        for (int i = 0; i < intervals.size(); i++) {
            Interval val = intervals.get(i);
            list.add(new Point(val.start, 1));
            list.add(new Point(val.end, 0));
        }
        Collections.sort(list, cmp);
        int count = 0;
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i).flag == 1) {
                count++;
            } else {
                count--;
            }
            if (count > 1) {
                return false;
            }
        }
        return true;
    }
}