题解 | #牛客直播各科目同时在线人数#

select course_id,course_name ,max(sum_num)
from (select distinct
           course_tb.course_id,
            course_name,
            sum(tag) over(partition by course_id order by datetime) as sum_num
        from course_tb
        left join (select
            user_id,
            course_id,
            in_datetime as datetime,
            1 as tag
        from attend_tb
        union
        select
            user_id,
            course_id,
            out_datetime as datetime,
            -1 as tag
        from attend_tb) main on main.course_id = course_tb.course_id) p group by course_id,course_name order  by course_id

没太明白这种思路。。。