题解 | #牛客直播各科目同时在线人数#

-又是一道求同时在线人数，老生常谈啦！还是那个套路，这就不多解释了

with t as(
select 
	a.course_id,
    course_name,
    in_datetime,
    out_datetime
from attend_tb a
left join course_tb b
on a.course_id=b.course_id)

select
	course_id,
    course_name,
    max(num) as max_num
from
(select 
course_id,
course_name,
sum(diff) over (partition by course_id order by dt,diff desc) num
from
(
select 
	course_id,
    course_name,
    in_datetime dt,
    1 diff
from t
union all
select 
	course_id,
    course_name,
    out_datetime dt,
    -1 diff
from t)t1)t2
group by course_id,course_name
order by course_id