feature:[MySQL].185.department top three salaries

Author: xrksudy

Diff for: README.md

@@ -45,6 +45,7 @@ LeetCode
 | 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)                  |
+| 185 | [Department Top Three Salaries](https://leetcode.com/problems/department-top-three-salaries/description/)                           | Hard       | [MySQL](./database/185.department-top-three-salaries.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)                    |
 

Diff for: database/185.department-top-three-salaries.sql

@@ -0,0 +1,79 @@
+--
+-- @lc app=leetcode id=185 lang=mysql
+--
+-- [185] Department Top Three Salaries
+--
+-- https://leetcode.com/problems/department-top-three-salaries/description/
+--
+-- database
+-- Hard (33.37%)
+-- Likes:    570
+-- Dislikes: 131
+-- Total Accepted:    79.8K
+-- Total Submissions: 230.5K
+-- Testcase Example:  '{"headers": {"Employee": ["Id", "Name", "Salary", "DepartmentId"], "Department": ["Id", "Name"]}, "rows": {"Employee": [[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": [[1, "IT"], [2, "Sales"]]}}'
+--
+-- The Employee table holds all employees. Every employee has an Id, and there
+-- is also a column for the department Id.
+-- 
+-- 
+-- +----+-------+--------+--------------+
+-- | 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            |
+-- +----+-------+--------+--------------+
+-- 
+-- 
+-- The Department table holds all departments of the company.
+-- 
+-- 
+-- +----+----------+
+-- | Id | Name     |
+-- +----+----------+
+-- | 1  | IT       |
+-- | 2  | Sales    |
+-- +----+----------+
+-- 
+-- 
+-- Write a SQL query to find employees who earn the top three salaries in each
+-- of the department. 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         | Randy    | 85000  |
+-- | IT         | Joe      | 85000  |
+-- | IT         | Will     | 70000  |
+-- | Sales      | Henry    | 80000  |
+-- | Sales      | Sam      | 60000  |
+-- +------------+----------+--------+
+-- 
+-- 
+-- Explanation:
+-- 
+-- In IT department, Max earns the highest salary, both Randy and Joe earn the
+-- second highest salary, and Will earns the third highest salary. There are
+-- only two employees in the Sales department, Henry earns the highest salary
+-- while Sam earns the second highest salary.
+-- 
+--
+
+-- @lc code=start
+-- Write your MySQL query statement below
+
+SELECT d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
+FROM Employee e1 JOIN Department d ON e1.DepartmentId = d.Id
+WHERE 3 > (SELECT COUNT(DISTINCT e2.Salary)
+FROM Employee e2
+WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId )
+
+-- @lc code=end