全站第 23/1430 名
解决了 83/335 题
中等: 17/76
入门: 17/77
困难: 3/29
简单: 40/114
草履虫: 6/39
过去1年一共提交 132 次
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勋章 ①金银铜:在竞赛中获得第一二三名;②好习惯:自然月10天提交;③里程碑:解决1/2/5/10/20/50/100/200题;④每周打卡挑战:完成每周5题,每年1月1日清零。
提交天数:14 天2025-06-30
达成1题里程碑2025-06-01
达成2题里程碑2025-06-01
达成5题里程碑2025-06-01
达成10题里程碑2025-06-10
达成20题里程碑2025-06-19
达成50题里程碑2025-06-30
完成w10打卡2025-07-02
完成w12打卡2025-06-10
完成w13打卡2025-06-01
完成w29打卡2025-06-28
完成w31打卡2025-06-25
完成w32打卡2025-06-03
完成w33打卡2025-06-03
完成w35打卡2025-07-02
完成w36打卡2025-06-03
完成w37打卡2025-06-30
完成w39打卡2025-06-10
完成w40打卡2025-06-01
完成w42打卡2025-06-03
完成w43打卡2025-06-01收藏
| 收藏日期 | 题目名称 | 解决状态 |
|---|---|---|
| 2025-06-26 | 基于共同兴趣爱好的餐厅推荐(5)-好基友(3) | 未解决 |
| 2025-06-19 | 找出所有以酒店为起点的类别组合的最热门路线 | 未解决 |
| 2025-06-01 | CASE WHEN-男女学生的数量 | 未解决 |
评论笔记
| 评论日期 | 题目名称 | 评论内容 | 站长评论 |
|---|---|---|---|
| 2025-06-11 | 给商品打四类标签(列) |
提交记录
| 提交日期 | 题目名称 | 提交代码 |
|---|---|---|
| 2025-08-07 | 条件过滤(2)半夜活动有猫腻,Hour函数给给力 |
select * from cmb_usr_trx_rcd where date(trx_time)between'2024-09-01'and'2024-09-30' and hour(trx_time)between 1 and 5 and usr_id='5201314520' |
| 2025-07-04 | 只买iPhone的用户 |
select user_id from apple_pchs_rcd group by user_id having sum(case when product_type != 'iphone' then 1 else 0 end) =0 order by 1 |
| 2025-07-04 | 至少两门科目大于等于110分的学生 |
select * from subject_score where (chinese>=110 and math >= 110) or (chinese>=110 and english >= 110) or (english >= 110 and math >= 110) order by student_id |
| 2025-07-03 | 抖音面试真题(5)新用户的T+1月留存 |
WITH monthly_unique_logins AS ( SELECT usr_id, DATE_FORMAT(login_time, '%Y-%m-01') AS login_month FROM user_login_log WHERE login_time >= '2024-01-01' AND login_time < '2025-01-01' GROUP BY usr_id, DATE_FORMAT(login_time, '%Y-%m-01') ), new_users AS ( SELECT usr_id, MIN(login_month) AS first_login_month FROM monthly_unique_logins GROUP BY usr_id ), next_month_logins AS ( SELECT nu.usr_id, nu.first_login_month AS current_month, mul.login_month AS next_month FROM new_users nu LEFT JOIN monthly_unique_logins mul ON nu.usr_id = mul.usr_id AND mul.login_month = DATE_ADD(nu.first_login_month, INTERVAL 1 MONTH) ) SELECT current_month, ROUND(COUNT(DISTINCT CASE WHEN next_month IS NOT NULL THEN usr_id END) * 100.0 / COUNT(DISTINCT usr_id), 2) AS t_plus_1_month_retention_rate FROM next_month_logins WHERE current_month >= '2024-01-01' AND current_month < '2025-01-01' GROUP BY current_month ORDER BY current_month; |
| 2025-07-03 | 1989年12月13日出生的女歌手 |
select * from singer_info where birth_date ='1989-12-13' |
| 2025-07-03 | 找出所有港台歌手 |
select * from singer_info where type2='港台' order by 1 |
| 2025-07-03 | 找出所有港台歌手 |
select * from singer_info where type2='港台' AND type3='个人' |
| 2025-07-03 | 找出所有港台乐队 |
select * from singer_info where type2='港台' and type3='乐队' order by singer_id |
| 2025-07-03 | 找出所有港台乐队 |
select * from singer_info where type2='港台' order by singer_id |
| 2025-07-03 | 能组成直角三角形的线段 |
SELECT * FROM numbers_for_fun WHERE CASE WHEN ABS(a) >= ABS(b) AND ABS(a) >= ABS(c) THEN POW(ABS(a), 2) = POW(ABS(b), 2) + POW(ABS(c), 2) WHEN ABS(b) >= ABS(a) AND ABS(b) >= ABS(c) THEN POW(ABS(b), 2) = POW(ABS(a), 2) + POW(ABS(c), 2) ELSE POW(ABS(c), 2) = POW(ABS(a), 2) + POW(ABS(b), 2) END and a*b*c<>0 ORDER BY id ASC; |
| 2025-07-03 | 能组成三角形的线段 |
select * from numbers_for_fun where a*b*c <> 0 and( abs(a)+abs(b)>abs(c) and abs(a)+abs(c)>abs(b) and abs(b)+abs(c)>abs(a) ) order by id; |
| 2025-07-03 | 大于J小于K的手牌 |
select * from hand_permutations where card1 between 'J' and'K' and card2 between 'J' and 'K' order by id ; |
| 2025-07-03 | 语文数学英语至少1门超过100分的同学 |
select * from subject_score where chinese >100 or math >100 or english >100 group by student_id order by chinese asc |
| 2025-07-03 | 统计每个作者发布视频的平均互动指数 |
WITH VideoInteractions AS ( SELECT v.author_id, SUM( w.if_like + CASE WHEN w.comment_id IS NOT NULL THEN 1 ELSE 0 END + w.if_retweet + w.if_fav ) AS total_interactions, COUNT(DISTINCT v.video_id) AS video_count FROM ks_video_inf v JOIN ks_video_wat_log w ON v.video_id = w.video_id GROUP BY v.author_id ) SELECT author_id, round(total_interactions * 1.0 / video_count,2) AS avg_interaction_index FROM VideoInteractions ORDER BY avg_interaction_index DESC; |
| 2025-07-03 | 计算视频的平均观看完成率 |
SELECT v.video_id, v.title, COALESCE(AVG(TIMESTAMPDIFF(SECOND, w.start_time, w.end_time) / v.duration), 0) AS avg_completion_rate FROM ks_video_inf v inner JOIN ks_video_wat_log w ON v.video_id = w.video_id GROUP BY v.video_id, v.title ORDER BY avg_completion_rate DESC; |
| 2025-07-03 | 找出最近一周内发布的竖屏视频 |
select video_id,author_id,title from ks_video_inf wherescreen_type = 'p' and date(release_time) >= date_sub(curdate(),interval 7 day) order by release_time desc |
| 2025-07-03 | 总分超过300分的学生 |
with B1 AS ( select student_id, chinese+math+english as cnt from subject_score) select t1.student_id from B1 t1 where cnt >= 300 |
| 2025-07-02 | 曝光量最大的商品 |
with B1 as ( select t1.cust_uid,t1.if_snd,t1.prd_id,t2.prd_nm,t2.price from tb_pg_act_rcd t1 left join tb_prd_map t2 on t1.prd_id=t2.prd_id ) select prd_id,prd_nm,sum(if_snd)as cnt from B1 group by prd_id,prd_nm order by sum(if_snd) desc limit 1; |
| 2025-07-02 | 登录天数分布 |
with B1 as (select usr_id, date(login_time) as day from user_login_log where login_time >= date_sub(curdate(),interval 180 day) group by usr_id,date(login_time) ), B2 as(select t1.usr_id, count(distinct t1.day) as cnt from B1 t1 group by t1.usr_id ), B3 as (select t2.usr_id, sum(case when t2.cnt BETWEEN 1 AND 5 then 1 else 0 end) as days_1_to_5, sum(case when t2.cnt between 6 and 10 then 1 else 0 end) as days_6_to_10, sum(case when t2.cnt between 11 and 20 then 1 else 0 end) as days_11_to_20, SUM(CASE WHEN t2.cnt > 20 THEN 1 ELSE 0 END) AS days_over_20 FROM B2 t2 group by 1) select sum(t3.days_1_to_5) as days_1_to_5_a, sum(t3.days_6_to_10) as days_6_to_10_a, sum(t3.days_11_to_20) as days_11_to_20_a, sum(t3.days_over_20) as days_over_20_a from B3 t3 ; |
| 2025-07-02 | 登录天数分布 |
with B1 as (select usr_id, date(login_time) as day from user_login_log group by usr_id,date(login_time) ), B2 as(select t1.usr_id, count(distinct t1.day) as cnt from B1 t1 group by t1.usr_id ), B3 as (select t2.usr_id, sum(case when t2.cnt <= 5 then 1 else 0 end) as days_1_to_5, sum(case when t2.cnt between 6 and 10 then 1 else 0 end) as days_6_to_10, sum(case when t2.cnt between 11 and 20 then 1 else 0 end) as days_11_to_20, SUM(CASE WHEN t2.cnt > 20 THEN 1 ELSE 0 END) AS days_over_20 FROM B2 t2 group by 1) select sum(t3.days_1_to_5) as days_1_to_5_a, sum(t3.days_6_to_10) as days_6_to_10_a, sum(t3.days_11_to_20) as days_11_to_20_a, sum(t3.days_over_20) as days_over_20_a from B3 t3 ; |