题解 | #获得积分最多的人(二)#

select
    distinct(id),
    name,
    grade_num
from (
        select
            id,
            name,
            grade_num,
            max(grade_num) over() as max_num    
        from
            (
                select
                    id,
                    name,
                    sum(grade_num) over(partition by user_id order by user_id) as grade_num           
                from
                    user u,
                    grade_info g
                where
                    u.id = g.user_id
            ) a

    ) b
where 
    b.grade_num = b.max_num