185. 部门工资前三高的员工

题目描述

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

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 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，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

题目难度：困难

分析

工资前三高的员工可以通过连表求出来。
连表的条件：e1.DepartmentId = e2.DepartmentId and e1.Salary <= e2.Salary
再通过group by e1.Id having count(distinct e2.Salary) <= 3 求得前三高的员工
部门的名字也是通过连表求出来
连表的条件:e1.DepartmentId = d.Id
最后输出结果需要进行排序
order by d.Name, e1.Salary desc;

代码

# Write your MySQL query statement below
select 
    d.Name Department, e1.Name Employee, e1.Salary
from
    Employee e1, Employee e2, Department d    
where
    e1.DepartmentId = e2.DepartmentId 
and
    e1.Salary <= e2.Salary
and
    e1.DepartmentId = d.Id
group by
    e1.Id    
having 
    count(distinct e2.Salary) <= 3   
order by
    d.Name, e1.Salary desc    
;

题目链接

https://leetcode-cn.com/problems/department-top-three-salaries/description/