Leetcode 252 meeting rooms

最新推荐文章于 2025-05-01 18:13:28 发布

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最新推荐文章于 2025-05-01 18:13:28 发布

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文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/u014748614/article/details/51726206

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本文介绍了一种用于检测一系列会议时间是否存在冲突的算法。通过将会议区间按开始时间排序并检查相邻会议的时间重叠情况来判断一个人是否可以参加所有预定的会议。此算法适用于日程安排软件等场景。

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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.

/**
 * 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; }
 * }
 */
public class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort( intervals,
            new Comparator<Interval> () {
                public int compare(Interval itv1, Interval itv2) {
                    return itv1.start - itv2.start;
                }
            }
            
        );
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start < intervals[i-1].end) {
                return false;
            }
        }
        return true;
    }
}

Best answer:

private boolean canAttendMeetings(Interval[] intervals) {
    try {
        Arrays.sort(intervals, new IntervalComparator());
    } catch (Exception e) {
        return false;
    }
    return true;
}

private class IntervalComparator implements Comparator<Interval> {
    @Override
    public int compare(Interval o1, Interval o2) {
        if (o1.start < o2.start && o1.end <= o2.start)
            return -1;
        else if (o1.start > o2.start && o1.start >= o2.end)
            return 1;
        throw new RuntimeException();
    }
}