Leetcode的SQL题解:185. 部门工资前三高的员工
题目
查询部门工资前三高的员工。
我用的数据库是oracle。
下面是数据表的信息。
Employee表数据:
| ID | NAME | Salary | DepartmentId |
| -- | ---- | ------ | ------------ |
|1 | Joe | 85000 | 1 |
|2 | Henry | 80000 | 2 |
|3 | Sam | 60000 | 2 |
|4 | Max | 90000 | 1 |
|5 | Janet | 69000 | 1 |
|6 | Randy | 85000 | 1 |
|7 | Will | 70000 | 1 |
|8 | edav | 50000 | 2 |
|9 | easonv | 40000 | 2 |
8、9行为我自行添加,为了更清晰展示查询结果。
创建表
Employee 表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id 。
create table Employee (
Id number(5),
Name varchar2(10) ,
Salary number(5),
DepartmentId number(5)
);
Department 表包含公司所有部门的信息。
create table Department (
Id number(5),
Name varchar2(10)
);
插入数据Employee,脚本如下
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('1', 'Joe', '85000', '1');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('2', 'Henry', '80000', '2');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('3', 'Sam', '60000', null);
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('4', 'Max', '90000', '1');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('5', 'Janet', '69000', '1');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('6', 'Randy', '85000', '1');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('7', 'Will', '70000', '1');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('8', 'eda', '50000', '2');
insert into Employee (ID, NAME, SALARY, DEPARTMENTID)
values ('9', 'eason', '40000', '2');
插入数据Department,脚本如下
insert into Department (ID, NAME)
values ('1', 'IT');
insert into Department (ID, NAME)
values ('2', 'Sales');
查询
以下使用四种SQL语句查出的结果,前两个是用oracle特有函数,后两个是标准SQL92写法。
你觉得哪个对?哪个性能高?
函数1 ROW_NUMBER
select Department,Employee,Salary
from (select (ROW_NUMBER()
over(PARTITION by t1.departmentid order by Salary desc)) lev,
t2.name Department,
t1.name Employee,
t1.Salary Salary
from Employee t1, Department t2
where t1.departmentid = t2.id) A
where lev <= 3;
函数2 dense_rank
select D.Name Department, E.Name Employee, E.Salary Salary
from (select Name,
Salary,
DepartmentId,
dense_rank() over(partition by DepartmentId order by Salary desc) Trank
from Employee) E
right join Department D
on E.DepartmentId = D.id
where Trank <= 3;
通用写法1
select d.name as Department, e.name as Employee, e.salary as Salary
from employee e
inner join department d
on e.DepartmentId = d.id
where (select count(distinct salary)
from employee
where salary > e.salary
and departmentid = e.DepartmentId) < 3
order by e.departmentid, Salary desc;
通用写法2
SELECT t3.name Department, t2.name Employee, t2.salary Salary
FROM Employee t2, Department t3
WHERE t2.id NOT IN (SELECT b.id
FROM Employee a, Employee b
WHERE a.DepartmentId = b.DepartmentId
AND a.salary > b.salary
GROUP BY b.id
HAVING COUNT(*) >= 3)
AND t2.DepartmentId = t3.id
ORDER BY Department, t2.salary DESC;
吐槽
感兴趣的同学可以自己跑下。
我个人觉得所谓官方答案是有问题的。
官方题解如下,mysql版本:
SELECT d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
FROM Employee e1
JOIN Department d
ON e1.DepartmentId = d.Id
WHERE 3 > (SELECT COUNT(DISTINCT e2.Salary)
FROM Employee e2
WHERE e2.Salary > e1.Salary
AND e1.DepartmentId = e2.DepartmentId);
改写成oracle版,加上排序:
SELECT d.Name Department, e1.Name Employee, e1.Salary
FROM Employee e1
JOIN Department d
ON e1.DepartmentId = d.Id
WHERE 3 > (SELECT COUNT(DISTINCT e2.Salary)
FROM Employee e2
WHERE e2.Salary > e1.Salary
AND e1.DepartmentId = e2.DepartmentId)
order by d.id,salary desc
查出来的数据是与通用写法1一样的,
两个同样的85000的数据
|序号| Department | Employee | Salary |
|--- | ---- | ------- | ------------ |
|1 | IT | Max | 90000
|2 | IT | Randy | 85000
|3 | IT | Joe | 85000
|4 | IT | Will | 70000
|5 | Sales | Henry | 80000
|6 | Sales | Sam | 60000
|7 | Sales | eda| 50000
这个题目出的歧义太大,如果是在考试中,应该是查出前三名、前四名的都对。
个人认为应该查出前三名应该是不包含70000这条数据的,就算是并列第二,那么就应该没有第三了,高校排名不也是这样吗?
所以私以为正确答案应该是查出这样的数据
|序号| Department | Employee | Salary |
|--- | ---- | ------- | ------------ |
|1 | IT | Max | 90000
|2 | IT | Randy | 85000
|3 | IT | Joe | 85000
|4 | Sales | Henry | 80000
|5 | Sales | Sam | 60000
|6 | Sales | eda| 50000
那么,我写的四条语句中,应该是函数1及通用写法2可以满足这个条件。
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