题解 | #每篇文章同一时刻最大在看人数#

记录一下这种解题思路，学到新东西了。

主要是不知道如何判断一个时间点的同时在线人数，一开始考虑了用单个时间点对每个uid的in_time和on_time都做差看正负，但是这样逻辑有点乱。看了题解的大佬思路，用1和-1代表进入和推出，然后通过时间排序后进行累计加减就能完成了，我得记住哈哈哈

select
k.artical_id, max(k.sum_wp) max_uv
from
    (
    select
    *, sum(t.diff) over(partition by t.artical_id order by wtime, t.diff desc) sum_wp
    from(
        select artical_id, in_time wtime, 1 diff
        from tb_user_log
        union all
        select artical_id, out_time wtime, -1 diff
        from tb_user_log
        ) t
        where t.artical_id <> 0
    ) k
group by k.artical_id
order by max_uv desc