LeetCode MySQL 185. 部门工资前三高的所有员工（dense_rank）

+----+-------+--------+--------------+
| 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            |
+----+-------+--------+--------------+

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

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

IT 部门中，
Max 获得了最高的工资，
Randy 和 Joe 都拿到了第二高的工资，
Will 的工资排第三。

销售部门（Sales）只有两名员工，
Henry 的工资最高，Sam 的工资排第二。

# Write your MySQL query statement below
select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e left join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3 and Department is not null

# Write your MySQL query statement below
select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e right join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3 and Employee is not null

select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3