老年人力扣SQL刷题(开通会员篇)之中位数
512. 游戏玩法分析 II
Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-05-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) 是这个表的两个主键
这个表显示的是某些游戏玩家的游戏活动情况
每一行是在某天使用某个设备登出之前登录并玩多个游戏（可能为0）的玩家的记录
请编写一个 SQL 查询，描述每一个玩家首次登陆的设备名称
查询结果格式在以下示例中：
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-----------+
| player_id | device_id |
+-----------+-----------+
| 1 | 2 |
| 2 | 3 |
| 3 | 1 |
+-----------+-----------+
题解一：子查询排序+关联
select activity.player_id, activity.device_id from activity, (
select player_id, min(event_date) as first_login
from activity group by player_id
) as temp
where activity.player_id = temp.player_id
and activity.event_date = temp.first_login;
题解二
select player_id, device_id from (
select
player_id,
device_id,
dense_rank() over(partition by player_id order by event_date) as 排名
from activity
) as temp where temp.排名 = 1;
题解三
select player_id, device_id from activity a1 where event_date <= all(
select a2.event_date from activity a2 where a1.player_id = a2.player_id
);
534. 游戏玩法分析 III
Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-05-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
（player_id，event_date）是此表的主键。
这张表显示了某些游戏的玩家的活动情况。
每一行是一个玩家的记录，他在某一天使用某个设备注销之前登录并玩了很多游戏（可能是 0 ）。
编写一个 SQL 查询，同时报告每组玩家和日期，以及玩家到目前为止玩了多少游戏。也就是说，在此日期之前玩家所玩的游戏总数。详细情况请查看示例。
查询结果格式如下所示：
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 1 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+------------+---------------------+
| player_id | event_date | games_played_so_far |
+-----------+------------+---------------------+
| 1 | 2016-03-01 | 5 |
| 1 | 2016-05-02 | 11 |
| 1 | 2017-06-25 | 12 |
| 3 | 2016-03-02 | 0 |
| 3 | 2018-07-03 | 5 |
+-----------+------------+---------------------+
对于 ID 为 1 的玩家，2016-05-02 共玩了 5+6=11 个游戏，2017-06-25 共玩了 5+6+1=12 个游戏。
对于 ID 为 3 的玩家，2018-07-03 共玩了 0+5=5 个游戏。
请注意，对于每个玩家，我们只关心玩家的登录日期。
题解一：
# Write your MySQL query statement below
SELECT player_id,event_date, SUM(games_played) OVER (partition by player_id order by event_date asc) AS games_played_so_far FROM Activity;
题解二
select a.player_id, a.event_date, sum(b.games_played) as games_played_so_far
from activity a join activity b
on a.player_id = b.player_id where a.event_date >= b.event_date
group by a.player_id, a.event_date;
550. 游戏玩法分析 IV
Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
（player_id，event_date）是此表的主键。
这张表显示了某些游戏的玩家的活动情况。
每一行是一个玩家的记录，他在某一天使用某个设备注销之前登录并玩了很多游戏（可能是 0）。
编写一个 SQL 查询，报告在首次登录的第二天再次登录的玩家的比率，四舍五入到小数点后两位。换句话说，您需要计算从首次登录日期开始至少连续两天登录的玩家的数量，然后除以玩家总数。
查询结果格式如下所示：
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
只有 ID 为 1 的玩家在第一天登录后才重新登录，所以答案是 1/3 = 0.33
题解一
select round(avg(a.event_date is not null), 2) fraction
from
(select player_id, min(event_date) as login
from activity
group by player_id) p
left join activity a
on p.player_id=a.player_id and datediff(a.event_date, p.login)=1
题解二
select round((
(select count(player_id) from (
select
player_id,
datediff(event_date, min(event_date) over(partition by player_id)) as diff
from activity
) as temp where diff = 1) / (select count(distinct player_id) from activity)
), 2) as fraction;
题解三
with temp as (
select
player_id,
datediff(event_date, min(event_date) over(partition by player_id)) as diff
from activity
) select round(
sum(case diff when 1 then 1 else 0 end) /
count(distinct player_id),
2) as fraction from temp;
题解四
select round(avg(event_date is not null), 2) as fraction from (
select player_id, min(event_date) as first_login from activity
group by player_id
) temp left join activity
on temp.player_id = activity.player_id
and datediff(event_date, first_login) = 1;
569. 员工薪水中位数
Create table If Not Exists Employee (id int, company varchar(255), salary int)
Truncate table Employee
insert into Employee (id, company, salary) values ('1', 'A', '2341')
insert into Employee (id, company, salary) values ('2', 'A', '341')
insert into Employee (id, company, salary) values ('3', 'A', '15')
insert into Employee (id, company, salary) values ('4', 'A', '15314')
insert into Employee (id, company, salary) values ('5', 'A', '451')
insert into Employee (id, company, salary) values ('6', 'A', '513')
insert into Employee (id, company, salary) values ('7', 'B', '15')
insert into Employee (id, company, salary) values ('8', 'B', '13')
insert into Employee (id, company, salary) values ('9', 'B', '1154')
insert into Employee (id, company, salary) values ('10', 'B', '1345')
insert into Employee (id, company, salary) values ('11', 'B', '1221')
insert into Employee (id, company, salary) values ('12', 'B', '234')
insert into Employee (id, company, salary) values ('13', 'C', '2345')
insert into Employee (id, company, salary) values ('14', 'C', '2645')
insert into Employee (id, company, salary) values ('15', 'C', '2645')
insert into Employee (id, company, salary) values ('16', 'C', '2652')
insert into Employee (id, company, salary) values ('17', 'C', '65')
表: Employee
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| id | int |
| company | varchar |
| salary | int |
+--------------+---------+
Id是该表的主键列。
该表的每一行表示公司和一名员工的工资。
写一个SQL查询，找出每个公司的工资中位数。
以 任意顺序 返回结果表。
查询结果格式如下所示。
示例 1:
输入:
Employee 表:
+----+---------+--------+
| id | company | salary |
+----+---------+--------+
| 1 | A | 2341 |
| 2 | A | 341 |
| 3 | A | 15 |
| 4 | A | 15314 |
| 5 | A | 451 |
| 6 | A | 513 |
| 7 | B | 15 |
| 8 | B | 13 |
| 9 | B | 1154 |
| 10 | B | 1345 |
| 11 | B | 1221 |
| 12 | B | 234 |
| 13 | C | 2345 |
| 14 | C | 2645 |
| 15 | C | 2645 |
| 16 | C | 2652 |
| 17 | C | 65 |
+----+---------+--------+
输出:
+----+---------+--------+
| id | company | salary |
+----+---------+--------+
| 5 | A | 451 |
| 6 | A | 513 |
| 12 | B | 234 |
| 9 | B | 1154 |
| 14 | C | 2645 |
+----+---------+--------+
进阶: 你能在不使用任何内置函数或窗口函数的情况下解决它吗?
# Write your MySQL query statement below
#外边是需要id,compny,salary
#内部，id,compant,salary,row_num over 排名， count(id) total
#条件就是排名 In(floor(total +1)/2,floor(total +2)/2)
SELECT id,company,salary
FROM(
SELECT id,company,salary,row_number() OVER(partition by company order by salary) 排名,
COUNT(id) OVER(partition by company) total
FROM Employee
) AS t1
WHERE t1.排名 in (floor((total + 1) / 2),floor((total + 2) / 2));
570. 至少有5名直接下属的经理
Create table If Not Exists Employee (id int, name varchar(255), department varchar(255), managerId int)
Truncate table Employee
insert into Employee (id, name, department, managerId) values ('101', 'John', 'A', 'None')
insert into Employee (id, name, department, managerId) values ('102', 'Dan', 'A', '101')
insert into Employee (id, name, department, managerId) values ('103', 'James', 'A', '101')
insert into Employee (id, name, department, managerId) values ('104', 'Amy', 'A', '101')
insert into Employee (id, name, department, managerId) values ('105', 'Anne', 'A', '101')
insert into Employee (id, name, department, managerId) values ('106', 'Ron', 'B', '101')
表: Employee
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
| department | varchar |
| managerId | int |
+-------------+---------+
Id是该表的主键列。
该表的每一行都表示雇员的名字、他们的部门和他们的经理的id。
如果managerId为空，则该员工没有经理。
没有员工会成为自己的管理者。
编写一个SQL查询，查询至少有5名直接下属的经理 。
以 任意顺序 返回结果表。
查询结果格式如下所示。
示例 1:
输入:
Employee 表:
+-----+-------+------------+-----------+
| id | name | department | managerId |
+-----+-------+------------+-----------+
| 101 | John | A | None |
| 102 | Dan | A | 101 |
| 103 | James | A | 101 |
| 104 | Amy | A | 101 |
| 105 | Anne | A | 101 |
| 106 | Ron | B | 101 |
+-----+-------+------------+-----------+
输出:
+------+
| name |
+------+
| John |
+------+
题解一
select name from Employee where id in (select managerId from Employee group by managerId having count(managerId) >= 5) ;
题解二
# Write your MySQL query statement below
select name from Employee t1, (select managerId from Employee group by managerId having count(managerId) >= 5) as t2 where t1.id = t2.managerId ;
571. 给定数字的频率查询中位数
Create table If Not Exists Numbers (num int, frequency int)
Truncate table Numbers
insert into Numbers (num, frequency) values ('0', '7')
insert into Numbers (num, frequency) values ('1', '1')
insert into Numbers (num, frequency) values ('2', '3')
insert into Numbers (num, frequency) values ('3', '1')
Numbers 表：
+-------------+------+
| Column Name | Type |
+-------------+------+
| num | int |
| frequency | int |
+-------------+------+
num 是这张表的主键。这张表的每一行表示某个数字在该数据库中的出现频率。
中位数 是将数据样本中半数较高值和半数较低值分隔开的值。
编写一个 SQL 查询，解压 Numbers 表，报告数据库中所有数字的 中位数 。结果四舍五入至 一位小数 。
查询结果如下例所示。
输入：
Numbers 表：
+-----+-----------+
| num | frequency |
+-----+-----------+
| 0 | 7 |
| 1 | 1 |
| 2 | 3 |
| 3 | 1 |
+-----+-----------+
输出：
+--------+
| median |
+--------+
| 0.0 |
+--------+
解释：
如果解压这个 Numbers 表，可以得到 [0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3] ，所以中位数是 (0 + 0) / 2 = 0 。
题解一
# Write your MySQL query statement below
select avg(num) as median from (
select
num,
sum(frequency) over(order by num desc) desc_frequency,
sum(frequency) over(order by num asc) asc_frequency,
sum(frequency) over() total_frequency
from numbers
) as temp
where desc_frequency >= total_frequency / 2
and asc_frequency >= total_frequency / 2;
题解二
select avg(num) as median from (
select n1.num from numbers n1 join numbers n2 group by n1.num
having sum(if(n1.num >= n2.num, n2.frequency, 0)) >= sum(n2.frequency) / 2
and sum(if(n1.num <= n2.num, n2.frequency, 0)) >= sum(n2.frequency) / 2
) as temp;
574. 当选者
Create table If Not Exists Candidate (id int, name varchar(255))
Create table If Not Exists Vote (id int, candidateId int)
Truncate table Candidate
insert into Candidate (id, name) values ('1', 'A')
insert into Candidate (id, name) values ('2', 'B')
insert into Candidate (id, name) values ('3', 'C')
insert into Candidate (id, name) values ('4', 'D')
insert into Candidate (id, name) values ('5', 'E')
Truncate table Vote
insert into Vote (id, candidateId) values ('1', '2')
insert into Vote (id, candidateId) values ('2', '4')
insert into Vote (id, candidateId) values ('3', '3')
insert into Vote (id, candidateId) values ('4', '2')
insert into Vote (id, candidateId) values ('5', '5')
表: Candidate
+-------------+----------+
| Column Name | Type |
+-------------+----------+
| id | int |
| name | varchar |
+-------------+----------+
Id是该表的主键列。
该表的每一行都包含关于候选对象的id和名称的信息。
表: Vote
+-------------+------+
| Column Name | Type |
+-------------+------+
| id | int |
| candidateId | int |
+-------------+------+
Id是自动递增的主键。
candidateId是id来自Candidate表的外键。
该表的每一行决定了在选举中获得第i张选票的候选人。
编写一个SQL查询来报告获胜候选人的名字(即获得最多选票的候选人)。
生成测试用例以确保 只有一个候选人赢得选举。
查询结果格式如下所示。
示例 1:
输入:
Candidate table:
+----+------+
| id | name |
+----+------+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | E |
+----+------+
Vote table:
+----+-------------+
| id | candidateId |
+----+-------------+
| 1 | 2 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 5 |
+----+-------------+
输出:
+------+
| name |
+------+
| B |
+------+
解释:
候选人B有2票。候选人C、D、E各有1票。
获胜者是候选人B。
题解一
# Write your MySQL query statement below
SELECT name FROM Candidate WHERE id =
(SELECT candidateId FROM Vote GROUP BY candidateId ORDER BY count(candidateId) desc limit 1)
题解二
select name from candidate join vote on candidate.id = vote.candidateId
group by name order by count(name) desc limit 1;
577. 员工奖金
Create table If Not Exists Employee (empId int, name varchar(255), supervisor int, salary int)
Create table If Not Exists Bonus (empId int, bonus int)
Truncate table Employee
insert into Employee (empId, name, supervisor, salary) values ('3', 'Brad', 'None', '4000')
insert into Employee (empId, name, supervisor, salary) values ('1', 'John', '3', '1000')
insert into Employee (empId, name, supervisor, salary) values ('2', 'Dan', '3', '2000')
insert into Employee (empId, name, supervisor, salary) values ('4', 'Thomas', '3', '4000')
Truncate table Bonus
insert into Bonus (empId, bonus) values ('2', '500')
insert into Bonus (empId, bonus) values ('4', '2000')
选出所有 bonus < 1000 的员工的 name 及其 bonus。
Employee 表单
+-------+--------+-----------+--------+
| empId | name | supervisor| salary |
+-------+--------+-----------+--------+
| 1 | John | 3 | 1000 |
| 2 | Dan | 3 | 2000 |
| 3 | Brad | null | 4000 |
| 4 | Thomas | 3 | 4000 |
+-------+--------+-----------+--------+
empId 是这张表单的主关键字
Bonus 表单
+-------+-------+
| empId | bonus |
+-------+-------+
| 2 | 500 |
| 4 | 2000 |
+-------+-------+
empId 是这张表单的主关键字
输出示例：
+-------+-------+
| name | bonus |
+-------+-------+
| John | null |
| Dan | 500 |
| Brad | null |
+-------+-------+
题解 使用ifnull
# Write your MySQL query statement below
SELECT name,bonus FROM Employee e LEFT JOIN Bonus b ON e.empId=b.empId WHERE ifnull(bonus,0) < 1000;
578. 查询回答率最高的问题
Create table If Not Exists SurveyLog (id int, action varchar(255), question_id int, answer_id int, q_num int, timestamp int)
Truncate table SurveyLog
insert into SurveyLog (id, action, question_id, answer_id, q_num, timestamp) values ('5', 'show', '285', 'None', '1', '123')
insert into SurveyLog (id, action, question_id, answer_id, q_num, timestamp) values ('5', 'answer', '285', '124124', '1', '124')
insert into SurveyLog (id, action, question_id, answer_id, q_num, timestamp) values ('5', 'show', '369', 'None', '2', '125')
insert into SurveyLog (id, action, question_id, answer_id, q_num, timestamp) values ('5', 'skip', '369', 'None', '2', '126')
SurveyLog 表：
+-------------+------+
| Column Name | Type |
+-------------+------+
| id | int |
| action | ENUM |
| question_id | int |
| answer_id | int |
| q_num | int |
| timestamp | int |
+-------------+------+
这张表没有主键，其中可能包含重复项。
action 是一个 ENUM 数据，可以是 "show"、"answer" 或者 "skip" 。
这张表的每一行表示：ID = id 的用户对 question_id 的问题在 timestamp 时间进行了 action 操作。
如果用户对应的操作是 "answer" ，answer_id 将会是对应答案的 id ，否则，值为 null 。
q_num 是该问题在当前会话中的数字顺序。
回答率 是指：同一问题编号中回答次数占显示次数的比率。
编写一个 SQL 查询以报告 回答率 最高的问题。如果有多个问题具有相同的最大 回答率 ，返回 question_id 最小的那个。
查询结果如下例所示。
示例：
输入：
SurveyLog table:
+----+--------+-------------+-----------+-------+-----------+
| id | action | question_id | answer_id | q_num | timestamp |
+----+--------+-------------+-----------+-------+-----------+
| 5 | show | 285 | null | 1 | 123 |
| 5 | answer | 285 | 124124 | 1 | 124 |
| 5 | show | 369 | null | 2 | 125 |
| 5 | skip | 369 | null | 2 | 126 |
+----+--------+-------------+-----------+-------+-----------+
输出：
+------------+
| survey_log |
+------------+
| 285 |
+------------+
解释：
问题 285 显示 1 次、回答 1 次。回答率为 1.0 。
问题 369 显示 1 次、回答 0 次。回答率为 0.0 。
问题 285 回答率最高。
题解一
# Write your MySQL query statement below
select
question_id as survey_log
from
(select
a.*,
dense_rank() over(order by sum(action = 'answer') / sum(action = 'show') desc, question_id) as rnk
# 这里用dense_rank或者其他都可以，因为不存在平局的情况
from SurveyLog a
group by question_id) t
where rnk=1