leetcode-数据库：184. 部门工资最高的员工

港控/mmm° 2024-04-18 10:41 61阅读 0赞

### 题目如下： ###

`Employee` 表包含所有员工信息，每个员工有其对应的 `Id`, `salary` 和 `departmentId`。

+----+-------+--------+--------------+
    | Id | Name  | Salary | DepartmentId |
    +----+-------+--------+--------------+
    | 1  | Joe   | 70000  | 1            |
    | 2  | Henry | 80000  | 2            |
    | 3  | Sam   | 60000  | 2            |
    | 4  | Max   | 90000  | 1            |
    +----+-------+--------+--------------+

`Department` 表包含公司所有部门的信息。

+----+----------+
    | Id | Name     |
    +----+----------+
    | 1  | IT       |
    | 2  | Sales    |
    +----+----------+

编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

+------------+----------+--------+
    | Department | Employee | Salary |
    +------------+----------+--------+
    | IT         | Max      | 90000  |
    | Sales      | Henry    | 80000  |
    +------------+----------+--------+

##### 解题思路 #####

两种方式：

*  使用子查询进行过滤
 *  使用join进行合并

### 使用子查询进行过滤（执行用时: 134 ms） ###

select d.Name Department,e.Name Employee,e.Salary Salary 
    from Employee e,Department d 
    where 
    	e.DepartmentId=d.Id 
    and 
    	(e.DepartmentId,e.Salary) 			#返回两列
    		in (
    			select DepartmentId,max(Salary) from Employee group by DepartmentId
    			)

返回结果：

> \{“headers”:\[“Department”,“Employee”,“Salary”\],  
> “values”:\[\[“IT”,“Jim”,90000\],\[“Sales”,“Henry”,80000\],\[“IT”,“Max”,90000\]\]\}

### 使用join联表，再过滤（执行用时: 129 ms） ###

先使用join查询出每个部门最高工资的部门名称、薪水

select DepartmentId, 
            d.name as Name ,
            max(Salary) as Salary from Employee  e
        join Department d on d.Id = e.DepartmentId 
        group by DepartmentId

> \{“headers”:\[“DepartmentId”,“Name”,“Salary”\],  
> “values”:\[\[1,“IT”,90000\],\[2,“Sales”,80000\]\]\}

再与`Employee`表进行联表查询，得出员工姓名

select 
    t.Name as Department,
    e.Name as Employee, 
    e.Salary as Salary
    from(
        select DepartmentId, 
            d.name as Name ,
            max(Salary) as Salary from Employee  e
        join Department d on d.Id = e.DepartmentId 
        group by DepartmentId) as t		# 表t查出 [[1,"IT",90000],[2,"Sales",80000]]
    join Employee e
    on t.DepartmentId = e.DepartmentId and e.Salary = t.Salary		# 连接条件（过滤）

得出正确结果：

> \{“headers”:\[“Department”,“Employee”,“Salary”\],  
> “values”:\[\[“IT”,“Jim”,90000\],\[“Sales”,“Henry”,80000\],\[“IT”,“Max”,90000\]\]\}