Leetcode SQL会员题【吐血总结~~】第四天

目录

1173. 即时食物配送 I
1174. 即时食物配送 II
1193. 每月交易 I
1194. 锦标赛优胜者
1204. 最后一个能进入电梯的人
1205. 每月交易II
1211. 查询结果的质量和占比
1212. 查询球队积分
1225. 报告系统状态的连续日期
1241. 每个帖子的评论数
1251. 平均售价
1264. 页面推荐
1270. 向公司CEO汇报工作的所有人
1280. 学生们参加各科测试的次数
1285. 找到连续区间的开始和结束数字

1173. 即时食物配送 I

• 源代码：

  select round(sum(case when order_date = customer_pref_delivery_date then 1 else 0 end) / count(delivery_id) * 100, 2) as immediate_percentage
  from delivery
  

• 思路：sum() + case when，使用这种方法让这个题目简化了很多，这也是一种经验吧

1174. 即时食物配送 II

• 源代码：

  with first_order as (
      select delivery_id
      from delivery
      where (customer_id, order_date) in (
          select customer_id, min(order_date)
          from delivery
          group by customer_id
      )
  )
  select 
      round(
          sum(case when order_date=customer_pref_delivery_date and delivery_id in (select * from first_order) then 1 else 0 end)
          / (select count(*) from first_order) * 100
      , 2) as immediate_percentage
  from delivery
  

• 思路：理解需求很重要，这个题目就是要求首次下单中即时订单的占比，因此我们可以先求出首次下单有哪些，然后利用子查询判断既是即时订单又是首次下单的订单个数

1193. 每月交易 I

• 源代码：

  select
      left(trans_date, 7) as month, country,
      count(id) as trans_count, 
      count(case when state = 'approved' then 1 else null end) as approved_count,
      sum(amount) as trans_total_amount,
      sum(case when state = 'approved' then amount else 0 end) as approved_total_amount
  from transactions
  group by left(trans_date, 7), country
  

• 思路：首先我们得知道如何获取到字符串前面的日期，可以有两种方式 substr() / left()；然后弄清楚分组字段，然后进行聚合就行了

1194. 锦标赛优胜者

• 源代码：

  select group_id, id player_id
  from (
      select group_id, id, row_number() over(partition by group_id order by total_score desc, id) rk
      from (
          select id, sum(score) total_score
          from (
              select match_id, first_player id, first_score score
              from matches
              union all
              select match_id, second_player id, second_score score
              from matches
          ) t
          group by id
      ) t
      join players
      on t.id = players.player_id
  ) t
  where rk = 1
  

• 思路：根据题目说到的需要对选手的得分进行累计，也就是不管它作为第一个选手还是第二个选手都需要进行累计，所以我们可以将一张表拆分为两张表来进行sum，这样就可以求到每个选手累计的分数，然后再去求组内分数最高的选手

1204. 最后一个能进入电梯的人

• 源代码：

  select person_name
  from (
      select person_name, sum(weight) over(order by turn) sum_weight
      from queue
  ) t
  where sum_weight <= 1000
  order by sum_weight desc
  limit 1;
  

• 思路：对sum函数进行开窗，就可以得到累计的体重和，只要取出小于等于1000且最大的那个人，就是最后一个体重合适并进入电梯的人

1205. 每月交易II **

• 源代码：

  with base as (
      select 'approved' tag, country, substr(trans_date, 1, 7) month, amount
      from Transactions where state = 'approved'
      union all
      select 'chargeback' tag, t.country, substr(cb.trans_date, 1, 7) month, t.amount
      from Chargebacks cb join Transactions t on cb.trans_id = t.id
  )
  select month, country, 
         sum(if(tag = 'approved', 1, 0)) approved_count,
         sum(if(tag = 'approved', amount, 0)) approved_amount,
         sum(if(tag = 'chargeback', 1, 0)) chargeback_count,
         sum(if(tag = 'chargeback', amount, 0)) chargeback_amount
  from base 
  group by month, country;
  

• 思路： 先对数据打上标签，然后union all，再进行分组聚合；
• 重点学习一下本题的思路，我记得 求直播间最大连续在线人数 也是这个思路

1211. 查询结果的质量和占比

• 源代码：

  select 
      query_name, 
      round(avg(rating / position), 2) quality, 
      round(sum(case when rating < 3 then 1 else 0 end) / count(*) * 100, 2) poor_query_percentage 
  from queries
  group by query_name
  

• 思路：需求比较简单，没有特别不容易懂的逻辑

1212. 查询球队积分

• 源代码：

  select team_id, team_name, ifnull(score, 0) as num_points 
  from teams 
  left join (
      select id, sum(score) score
      from (
          -- 作为主场
          select  
              host_team id, 
              sum(case when host_goals > guest_goals then 3 when host_goals = guest_goals then 1 else 0 end) score
          from matches
          group by id
          union all 
          -- 作为客场
          select  
              guest_team id, 
              sum(case when host_goals < guest_goals then 3 when host_goals = guest_goals then 1 else 0 end) score
          from matches
          group by id
      ) t
      group by id
  ) t
  on teams.team_id = t.id
  order by num_points desc, team_id;
  

• 思路：核心逻辑就是left join右边的逻辑，我们首先计算主场的得分，然后计算客场的得分，最后将这两部分累计到一起

1225. 报告系统状态的连续日期 **

• 源代码：

  select type period_state , min(dt) start_date, max(dt) end_date
  from 
  (
      select  
          type, dt,
          subdate(dt, row_number() over(partition by type order by dt)) diff
      from (
          select 'succeeded' type, success_date dt
          from succeeded
          where success_date between '2019-01-01' and '2019-12-31'
          union 
          select 'failed' type, fail_date dt
          from failed
          where fail_date between '2019-01-01' and '2019-12-31'
      ) t
  ) t
  group by type, diff
  order by start_date;
  

• 思路：类似于求最大连续登陆天数问题，不同的是本题求 连续登陆的起止时间！这个题目需要对两张表进行连接，然后需要自己创建一个字段（想法非常好），然后再套用最大连续登陆问题套路date_sub + row_number来解决

1241. 每个帖子的评论数

• 源代码：

  select sub_id post_id, ifnull(number_of_comments, 0) number_of_comments
  from (
      select distinct sub_id
      from submissions
      where parent_id is null
  ) t1
  left join (
      select parent_id, count(distinct sub_id) number_of_comments 
      from submissions
      where parent_id is not null
      group by parent_id
  ) t2
  on t1.sub_id = t2.parent_id
  order by post_id;
  

• 思路：只要分清楚每一条数据是帖子，还是评论就可以了；然后还要注意去重！

1251. 平均售价

• 源代码：

  select 
      product_id, 
      round(sum(price * units) / sum(units), 2) average_price
  from 
  (
      select p.product_id, price, units 
      from prices p 
      join unitssold u 
      on p.product_id = u.product_id and purchase_date between start_date and end_date
  ) t
  group by product_id;
  

• 思路：比较简单

1264. 页面推荐

• 源代码：

  select distinct page_id as recommended_page
  from (
      select page_id 
      from likes
      where user_id in (
          select case when user1_id = 1 then user2_id else user1_id end 
          from friendship 
          where user1_id = 1 or user2_id = 1
      ) 
  ) t
  where page_id not in (
      select page_id
      from likes
      where user_id = 1
  )
  ;
  

• 思路：比较简单，需要对其中一些细节注意一下就可以了

1270. 向公司CEO汇报工作的所有人 *

• 源代码：

  select 
     e3.employee_id
  from employees e1, employees e2, employees e3 
  where 
     -- 根据题目说的要找出所有经理为1的员工id
     e1.manager_id  = 1 and 
     -- 根据上面一个条件找出直接经理为1的员工id，然后让这些员工id作为第二张表的经理，来获取所有的员工id
     e2.manager_id = e1.employee_id and  
     -- 根绝上面一个条件找出直接或者间接一次得到的所有员工id，然后让这些员工id作为第三张表的经理，来获取所有的员工id
     e3.manager_id = e2.employee_id and 
     -- 排除自己的经理就是自己的员工id
     e3.employee_id != 1
  ;
  

• 思路：这个题目如果不写程序很容易得到答案，但是通过程序实现就比较难了，我在上述源码中解释了 where后面的每一句话的意思

1280. 学生们参加各科测试的次数

• 源代码：

  select t1.student_id, student_name, t1.subject_name, count(ex.student_id) attended_exams
  from (
     select student_id, student_name, subject_name
     from students
     join subjects
     on 1 = 1
  ) t1
  left join examinations ex
  on t1.student_id = ex.student_id and t1.subject_name = ex.subject_name
  group by t1.student_id, t1.subject_name
  order by t1.student_id, t1.subject_name;
  

• 思路：其中涉及到一个笛卡尔积连接，可以使用cross join

1285. 找到连续区间的开始和结束数字

• 源代码：

  select 
     min(log_id) start_id,
     max(log_id) end_id
  from (
     select log_id, (log_id - row_number() over(order by log_id)) diff
     from logs
  ) t
  group by diff
  order by start_id
  

• 思路：和上面一个题一模一样，求连续登陆的最大时间和最小时间