# 部门工资前三高的所有员工 - LeetCode

Employee 表包含所有员工信息，每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。

+----+-------+--------+--------------+
| 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 表包含公司所有部门的信息。

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

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。

select
d.Name as 'Department',
e.Name as 'Employee',
e.Salary as 'Salary'
from Employee e
join Department d
on e.DepartmentId = d.Id
where 3 > (
select
count(distinct e2.Salary)
from Employee e2
where e2.DepartmentId = e.DepartmentId
and e2.Salary > e.Salary
)


MySQL解答：

# Write your MySQL query statement below

select
d.Name as Department,
temp.Name Employee,
temp.Salary
from
Department d left join
(select
e.DepartmentId,
e.Name,
@curRank := if (@prevDept = DepartmentId, if(@prevSal = e.Salary, @curRank, @curRank + 1), 1) as Rank,
@prevSal := e.Salary as Salary,
@prevDept := e.DepartmentId
from Employee e,
(select @prevDept := null, @curRank := 0, @prevSal := null) t
order by e.DepartmentId, e.Salary desc
) temp on d.Id = temp.DepartmentId
where temp.Rank <= 3

: )

posted @ 2019-11-29 17:14  dai.sp  阅读(794)  评论(0编辑  收藏  举报