题解 | #统计每个学校的答过题的用户的平均答题数#

写了两种方法，我想知道在实际工作中那种效率更高？求大牛讲解一下，谢谢啦！！

#方法一：
SELECT
    university
    ,ROUND(COUNT(q.result)/COUNT(DISTINCT q.device_id),4) avg_answer_cnt
FROM
    user_profile u
JOIN
    question_practice_detail q
WHERE
    u.device_id = q.device_id
GROUP BY 
    university;

#方法二：答案是一样的，只是牛客网的评改系统过不了
SELECT 
    t1.un university
    ,ROUND(SUM(c_re)/COUNT(t1.de_id),4) avg_answer_cnt
FROM(
    SELECT
        university un
        ,q.device_id de_id
        ,COUNT(result) c_re
    FROM
        user_profile u
    JOIN
        question_practice_detail q
    WHERE
        u.device_id = q.device_id
    GROUP BY 
        university
        ,q.device_id
    ) AS t1
GROUP BY
    t1.un