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【SQL】LeetCode-Department Highest Salary

技術標籤:SQL

LeetCode 184:Department Highest Salary

【Description】

The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.
+----+-------+--------+--------------+| Id | Name  | Salary | DepartmentId |+----+-------+--------+--------------+| 1  | Joe   | 70000  | 1            || 2  | Jim   | 90000  | 1            || 3  | Henry | 80000  | 2            || 4  | Sam   | 60000  | 2            || 5  | Max   | 90000  | 1            |+----+-------+--------+--------------+
The Department table holds all departments of the company.
+----+----------+| Id | Name     |+----+----------+| 1  | IT       || 2  | Sales    |+----+----------+
Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, your SQL query should return the following rows (order of rows does not matter).

+------------+----------+--------+| Department | Employee | Salary |+------------+----------+--------+| IT         | Max      | 90000  || IT         | Jim      | 90000  || Sales      | Henry    | 80000  |+------------+----------+--------+
Explanation:
Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

【Solution】

Personal solution:

with temp as (
    select *,
        rank() over (partition by DepartmentId
                     order by Salary desc) as "rank"
from Employee ) select d.Name as Department, e.Name as Employee, e.Salary as Salary from temp as e join Department as d on e.DepartmentId= d.Id where e.rank=1

Reference solution:

SELECT
    Department.name AS 'Department',
    Employee.name AS 'Employee',
    Salary
FROM
    Employee
        JOIN
Department ON Employee.DepartmentId = Department.Id WHERE (Employee.DepartmentId , Salary) IN ( SELECT DepartmentId, MAX(Salary) FROM Employee GROUP BY DepartmentId ) ;