leetcode184 部门工资最高的员工 Department Highest Salary

Employee表有全部员工。每一个员工有 Id,salary 和 department Id 信息。

 

 

建立表和数据：

 

 

drop table Employee
Create table If Not Exists Employee (Idint, Name varchar(255), Salary int, DepartmentId int);
drop table Department
Create table If Not Exists Department (Idint, Name varchar(255));
Truncate table Employee;
insert into Employee (Id, Name, Salary,DepartmentId) values ('1', 'Joe', '70000', '1');
insert into Employee (Id, Name, Salary,DepartmentId) values ('2', 'Henry', '80000', '2');
insert into Employee (Id, Name, Salary,DepartmentId) values ('3', 'Sam', '60000', '2');
insert into Employee (Id, Name, Salary,DepartmentId) values ('4', 'Max', '90000', '1');
Truncate table Department;
insert into Department (Id, Name) values('1', 'IT');
insert into Department (Id, Name) values('2', 'Sales');

 

解法：

1.先找出每一个部门的最高薪水。

链接员工表和部门表，group by对部门分组，再求每组的最高薪水。用子查询得出临时表F(id,name,m)。

(
    select D.id,D.name,max(E1.salary) as m
    from Department as D
    join Employee as E1 
        on (D.id = E1.departmentid)
    group by D.id,D.name
   ) as F

再次，链接员工表和临时表F，条件是部门id相同且薪水与每一个部门最高薪水相同。与上面的组合起来得出算法：

 

 

select F.name as Department,E.name as Employee,E.Salary
from Employee as E
join (
    select D.id,D.name,max(E1.salary) as m
    from Department as D
    join Employee as E1 
        on (D.id = E1.departmentid)
    group by D.id,D.name
) as F
    on (E.departmentid = F.id and F.m = E.salary);

员工表中有部门id和薪水，但没有部门name。可先求部门的最高薪水，造成临时表，再链接员工表和部门表，取出部门name和员工name。则有下面的算法：

select D.name as Department,E.name as Employee,E.Salary
from Employee as E
join (
    select E1.departmentid,max(E1.salary) as m
    from Employee as E1
    group by E1.departmentid
) as F
    on (E.departmentid = F.departmentid and F.m = E.salary)
join Department as D
    on (F.departmentid = D.id)

2.对每个员工，找出员工所在部门的最高薪水。此处的查找过程，用子查询实现。若是员工的薪水等于部门的最高薪水就是结果。

select D.name as Department,E.name as Employee,E.Salary
from Employee as E
join Department as D
    on (E.departmentid = D.id)
where E.salary = (
    select max(E1.salary) 
    from Employee as E1
    where E1.departmentid = E.departmentid
)

二元组(部门id,部门最高薪水)做为总体t，结合in判断t是否在已存在的组合中。

select D.name as Department,E.name as Employee,E.Salary
from Employee as E
join Department as D
    on (E.departmentid = D.id)
where (E.departmentid,E.salary) in (
    select E1.departmentid,max(E1.salary) 
    from Employee as E1
    where E1.departmentid
    group by E1.departmentid
)