with a as (select id, name, sum(grade_num)over(partition by user_id) as grade_sum from user u join grade_info g on u.id=g.user_id order by grade_sum desc ) select id, name, grade_sum from( select *, dense_rank()over(order by grade_sum desc) rk from a) t1 where rk=1 group by id order by id 大佬为啥这样会报错答案一样
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11-03 13:18
门头沟学院 Java
包行:平时怎么刷算法题的哇,字节的手撕听说都很难
字节跳动工作体验
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