运动场上有很多组照明设备，下面A集合中，描述出了不同照明设备的启动（begintime）停止（endtime）时间，照明设备的id已经处理掉了无需考虑，仔细看这些时间段可以发现，这些时间有重合（表示这段时间多组照明同时亮）、有空挡（表示这段时间一组照明都没有开）。请通过java代码实现，获得此足球场上有照明设备开启的时间段集合（求A集合在时间维度上的并集）。 A集合： [{begintime:2020-01-01 00:04:00,endtime:2020-01-01 00:06:00}, {begintime:2020-01-01 00:01:05,endtime:2020-01-01 00:03:15}, {begintime:2020-01-01 00:09:00,endtime:2020-01-01 00:10:00}, {begintime:2020-01-01 00:07:35,endtime:2020-01-01 00:08:40}, {begintime:2020-01-01 00:12:00,endtime:2020-01-01 00:17:00}, {begintime:2020-01-01 00:12:05,endtime:2020-01-01 00:13:10}, {begintime:2020-01-01 00:19:00,endtime:2020-01-01 00:21:00}, {begintime:2020-01-01 00:15:20,endtime:2020-01-01 00:20:45}]  期望得到的输出结果： [{begintime:2020-01-01 00:01:05,endtime:2020-01-01 00:03:15}, {begintime:2020-01-01 00:04:00,endtime:2020-01-01 00:06:00}, {begintime:2020-01-01 00:07:35,endtime:2020-01-01 00:08:40}, {begintime:2020-01-01 00:09:00,endtime:2020-01-01 00:10:00}, {begintime:2020-01-01 00:12:00,endtime:2020-01-01 00:21:00}]

可以使用如下的 Java 代码实现：

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.*;

public class Main {
    public static void main(String[] args) throws ParseException {
        List<Map<String, String>> A = new ArrayList<>();
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:04:00"); put("endtime", "2020-01-01 00:06:00");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:01:05"); put("endtime", "2020-01-01 00:03:15");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:09:00"); put("endtime", "2020-01-01 00:10:00");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:07:35"); put("endtime", "2020-01-01 00:08:40");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:12:00"); put("endtime", "2020-01-01 00:17:00");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:12:05"); put("endtime", "2020-01-01 00:13:10");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:19:00"); put("endtime", "2020-01-01 00:21:00");}});
        A.add(new HashMap<String, String>() {{put("begintime", "2020-01-01 00:15:20"); put("endtime", "2020-01-01 00:20:45");}});

        List<Map<String, String>> result = mergeTime(A);
        System.out.println(result);
    }

    public static List<Map<String, String>> mergeTime(List<Map<String, String>> A) throws ParseException {
        List<Map<String, String>> result = new ArrayList<>();
        if (A == null || A.size() == 0) {
            return result;
        }

        // 按照起始时间排序
        Collections.sort(A, new Comparator<Map<String, String>>() {
            public int compare(Map<String, String> o1, Map<String, String> o2) {
                SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
                try {
                    Date beginTime1 = sdf.parse(o1.get("begintime"));
                    Date beginTime2 = sdf.parse(o2.get("begintime"));
                    return beginTime1.compareTo(beginTime2);
                } catch (ParseException e) {
                    e.printStackTrace();
                    return 0;
                }
            }
        });

        String beginTime = A.get(0).get("begintime");
        String endTime = A.get(0).get("endtime");
        for (int i = 1; i < A.size(); i++) {
            String curBeginTime = A.get(i).get("begintime");
            String curEndTime = A.get(i).get("endtime");
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            Date endTimeDate = sdf.parse(endTime);
            Date curBeginTimeDate = sdf.parse(curBeginTime);

            if (endTimeDate.compareTo(curBeginTimeDate) >= 0) {  // 有重合
                endTime = curEndTime.compareTo(endTime) > 0 ? curEndTime : endTime;
            } else {  // 无重合
                result.add(new HashMap<String, String>() {{put("begintime", beginTime); put("endtime", endTime);}});
                beginTime = curBeginTime;
                endTime = curEndTime;
            }
        }

        result.add(new HashMap<String, String>() {{put("begintime", beginTime); put("endtime", endTime);}});
        return result;
    }
}

在程序中，首先按照起始时间对照明设备进行排序。然后遍历整个集合，对于当前照明设备，如果其起始时间与前一个照明设备的结束时间有重合，那么将它们合并为一个时间段；否则将前一个时间段加入结果集合中，并开始新的时间段。最后将最后一个时间段加入结果集合中即可。