题解 | 每篇文章同一时刻最大在看人数

思路：首先筛掉不符合的artical_id = 0 的情况，接着通过打标签的方式，将in_time和out_time转为一列，使用1和-1进行区分，使用sum开窗函数累加得到每次人数发生变化时的在线人数，求max，作order处理。
-- 筛 artical_id = 0
-- 打标签in 1 out -1
-- sum() over 按文章
-- desc
with t1 as (
   select
    artical_id,
    in_time,
    out_time
from tb_user_log
where artical_id != 0 
)
select
    artical_id,
    max(uv) as max_uv
from(
    select
    artical_id,
    sum(flag) over(partition by artical_id order by view_time asc, flag desc) uv
from(select
    artical_id,
    in_time as view_time,
    1 as flag
from t1
union all
select
    artical_id,
    out_time as view_time,
    -1 as flag
from t1)t2
)t3
group by artical_id
order by max_uv desc