高级资深攻城狮
高级资深攻城狮
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题解 | #数组中重复的数字#
随便写写,利用了set存储元素不重复的特性,在利用set的contains()函数,这样能把复杂度降到最低;
数组
2022-08-23
0
144
题解 | #分组过滤练习题#
select replace(SUBSTRING(start_time,1,7),"-",""), round(count(distinct uid, date_format(submit_time, '%Y%m%d')) / count(distinct uid),2), count(distin...
Mysql
2021-11-17
0
217
题解 | #分组过滤练习题#
select device_id,up1.university,up1.gpa from user_profile as up1,(select university,min(gpa)as gpa from user_profile group by university)as up2 where ...
Mysql
2021-10-20
0
191
题解 | #分组过滤练习题#
select device_id,up1.university,up1.gpa from user_profile as up1,(select university,min(gpa)as gpa from user_profile group by university)as up2 where ...
Mysql
2021-10-20
0
218
题解 | #分组过滤练习题#
select university,difficult_level,count(up.device_id)/ count(distinct(up.device_id)) from user_profile as up,question_practice_detail as qpd,question_...
Mysql
2021-10-18
0
187
题解 | #分组过滤练习题#
select university,count(question_id)/ count(distinct up.device_id) as avga from user_profile as up,question_practice_detail as qpd where up.device_id ...
Mysql
2021-10-15
0
190
题解 | #分组过滤练习题#
select qpd.device_id,question_id,result from question_practice_detail as qpd,user_profile as up where qpd.device_id=up.device_id and university="浙江大学"
Mysql
2021-10-15
0
226
题解 | #获得积分最多的人(二)#
select distinct music_name from follow as f,music_likes as ml,music as m where f.user_id='1' and f.follower_id=ml.user_id and ml.music_id=m.id and ml....
Mysql
2021-10-15
0
251
题解 | #获得积分最多的人(二)#
select user_id,name,sum from ( select user_id,name,sum(if(type='reduce',-1grade_num,grade_num))as sum, (select sum(if(type='reduce',-1grade_num,grade_...
Mysql
2021-10-15
0
193
题解 | #获得积分最多的人(二)#
select user_id,name,s from (select user_id,name,sum(grade_num) as s,m from grade_info as gi,user as u,(select sum(grade_num)as m from grade_info group...
Mysql
2021-10-15
0
211
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