此题为sql困难题，值得记录一下

题目描述

来自力扣第185题

输入: 
Employee 表:
+----+-------+--------+--------------+
| 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            |
+----+-------+--------+--------------+
Department  表:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
输出: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+
解释:
在IT部门:
- Max的工资最高
- 兰迪和乔都赚取第二高的独特的薪水
- 威尔的薪水是第三高的

在销售部:
- 亨利的工资最高
- 山姆的薪水第二高
- 没有第三高的工资，因为只有两名员工

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/department-top-three-salaries
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数据源

Create table If Not Exists Employee (id int, name varchar(255), salary int, departmentId int)
Create table If Not Exists Department (id int, name varchar(255))
Truncate table Employee
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', '2')
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')
Truncate table Department
insert into Department (id, name) values ('1', 'IT')
insert into Department (id, name) values ('2', 'Sales')

题解一

自联接，查询出(同一个部门id下，比每一个薪水大的个数)的集合，此个数小于3，即取出了前三

select D.name as Department, A.name as Employee, A.salary 
from Employee A, Department D
where 
(
    select count(0) 
    from (select distinct salary, departmentId from Employee) B 
    where departmentId = A.departmentId and salary > A.salary
) < 3 and A.departmentId = D.id
order by departmentId, salary desc

题解二

使用dense_rank函数直观的按部门分组，按薪水排序

select Department, Employee, Salary
from (
    select d.Name as Department, e.Name as Employee, e.Salary as Salary, 
dense_rank() over ( partition by DepartmentId order by Salary desc) as rk
    from Employee as e, Department as d
    where e.DepartmentId = d.Id
) m
where rk <= 3;