SQL面试题挑战02：同时最大在线人数问题

问题：

问题：如下为某直播平台各主播的开播及关播时间明细数据，现在需要计算出该平台最高峰期同时在线的主播人数。

user_id     start_date              end_date
1001    2021-06-14 12:12:12     2021-06-14 18:12:12
1003    2021-06-14 13:12:12     2021-06-14 16:12:12
1004    2021-06-14 13:15:12     2021-06-14 20:12:12
1002    2021-06-14 15:12:12     2021-06-14 16:12:12
1005    2021-06-14 15:18:12     2021-06-14 20:12:12
1001    2021-06-14 20:12:12     2021-06-14 23:12:12
1006    2021-06-14 21:12:12     2021-06-14 23:15:12
1007    2021-06-14 22:12:12     2021-06-14 23:10:12

SQL解答：

这是非常经典的一个面试题，不管大厂小厂都有问到过。解题思路也比较固定：就是用1代表开播(此时用开播时间)，-1代表关播(此时用关播时间)，可以理解1代表主播开播加入增1，-1代表主播关播离开减1，然后开窗可以计算出到每个时间点时有多少主播同时在线，最后求最大值即可。

with tmp as
(
    select 1001 as user_id, '2021-06-14 12:12:12' as start_date , '2021-06-14 18:12:12' as end_date
    union all
    select 1003 as user_id, '2021-06-14 13:12:12' as start_date , '2021-06-14 16:12:12' as end_date
    union all
    select 1004 as user_id, '2021-06-14 13:15:12' as start_date , '2021-06-14 20:12:12' as end_date
    union all
    select 1002 as user_id, '2021-06-14 15:12:12' as start_date , '2021-06-14 16:12:12' as end_date
    union all
    select 1005 as user_id, '2021-06-14 15:18:12' as start_date , '2021-06-14 20:12:12' as end_date
    union all
    select 1001 as user_id, '2021-06-14 20:12:12' as start_date , '2021-06-14 23:12:12' as end_date
    union all
    select 1006 as user_id, '2021-06-14 21:12:12' as start_date , '2021-06-14 23:15:12' as end_date
    union all
    select 1007 as user_id, '2021-06-14 22:12:12' as start_date , '2021-06-14 23:10:12' as end_date
)
select
max(online_nums) as max_online_nums
from
(
    select
    user_id
    ,dt
    ,sum(flag) over(order by dt) as online_nums
    from
    (
        select
        user_id
        ,start_date as dt
        ,1 as flag  --开播记为1
        from tmp
        union all
        select
        user_id
        ,end_date as dt
        ,-1 as flag --关播记为-1
        from tmp
    )t1
)t1
;