feature:[MySQL].184.department highest salary

Author: xrksudy

README.md

@@ -43,6 +43,7 @@ LeetCode
 | 181 | [Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers/description/) | Easy       | [MySQL](./database/181.employees-earning-more-than-their-managers.sql) |
 | 182 | [Duplicate Emails](https://leetcode.com/problems/duplicate-emails/description/)                                                     | Easy       | [MySQL](./database/182.duplicate-emails.sql)                           |
 | 183 | [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/)                                               | Easy       | [MySQL](./database/183.customers-who-never-order.sql)                  |
+| 184 | [Department Highest Salary](https://leetcode.com/problems/department-highest-salary/description/)                                   | Medium     | [MySQL](./database/184.department-highest-salary.sql)                  |
 | 192 | [Rising Temperature](https://leetcode.com/problems/rising-temperature/)                                                             | Easy       | [MySQL](./database/197.rising-temperature.sql)                         |
 | 196 | [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/description/)                                       | Easy       | [MySQL](./database/196.delete-duplicate-emails.sql)                    |
 

database/184.department-highest-salary.sql

@@ -0,0 +1,76 @@
+--
+-- @lc app=leetcode id=184 lang=mysql
+--
+-- [184] Department Highest Salary
+--
+-- https://leetcode.com/problems/department-highest-salary/description/
+--
+-- database
+-- Medium (35.76%)
+-- Likes:    480
+-- Dislikes: 112
+-- Total Accepted:    111.4K
+-- Total Submissions: 303K
+-- Testcase Example:  '{"headers": {"Employee": ["Id", "Name", "Salary", "DepartmentId"], "Department": ["Id", "Name"]}, "rows": {"Employee": [[1, "Joe", 70000, 1], [2, "Jim", 90000, 1], [3, "Henry", 80000, 2], [4, "Sam", 60000, 2], [5, "Max", 90000, 1]], "Department": [[1, "IT"], [2, "Sales"]]}}'
+--
+-- 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.
+-- 
+--
+
+-- @lc code=start
+-- Write your MySQL query statement below
+
+SELECT b.Name AS Department, a.Name AS Employee, Salary
+FROM Employee a JOIN Department b ON a.DepartmentId = b.Id
+WHERE (a.DepartmentId, a.Salary
+) IN
+(SELECT
+  DepartmentId, MAX(Salary)
+FROM
+  Employee
+GROUP BY DepartmentId)
+
+-- @lc code=end