SQL实现LeetCode(185.系里前三高薪水)

[LeetCode] 185.Department Top Three Salaries 系里前三高薪水

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   | 70000 | 1            |
| 2 | Henry | 80000 | 2            |
| 3 | Sam   | 60000 | 2            |
| 4 | Max   | 90000 | 1            |
| 5 | Janet | 69000 | 1            |
| 6 | Randy | 85000 | 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.

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

这道题是之前那道Department Highest Salary的拓展，难度标记为Hard，还是蛮有难度的一道题，综合了前面很多题的知识点，首先看使用Select Count(Distinct)的方法，我们内交Employee和Department两张表，然后我们找出比当前薪水高的最多只能有两个，那么前三高的都能被取出来了，参见代码如下：

解法一：

				?

									SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM Employee e

									JOIN Department d on e.DepartmentId = d.Id

									WHERE (SELECT COUNT(DISTINCT Salary) FROM Employee WHERE Salary > e.Salary

									AND DepartmentId = d.Id) < 3 ORDER BY d.Name, e.Salary DESC;

下面这种方法将上面方法中的<3换成了IN (0, 1, 2)，是一样的效果：

解法二：

				?

									SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM Employee e, Department d

									WHERE (SELECT COUNT(DISTINCT Salary) FROM Employee WHERE Salary > e.Salary

									AND DepartmentId = d.Id) IN (0, 1, 2) AND e.DepartmentId = d.Id ORDER BY d.Name, e.Salary DESC;

或者我们也可以使用Group by Having Count(Distinct ..) 关键字来做：

解法三：

				?

									SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM

									(SELECT e1.Name, e1.Salary, e1.DepartmentId FROM Employee e1 JOIN Employee e2 

									ON e1.DepartmentId = e2.DepartmentId AND e1.Salary <= e2.Salary GROUP BY e1.Id 

									HAVING COUNT(DISTINCT e2.Salary) <= 3) e JOIN Department d ON e.DepartmentId = d.Id 

									ORDER BY d.Name, e.Salary DESC;

下面这种方法略微复杂一些，用到了变量，跟Consecutive Numbers中的解法三使用的方法一样，目的是为了给每个人都按照薪水的高低增加一个rank，最后返回rank值小于等于3的项即可，参见代码如下：

解法四：

				?

									SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM

									(SELECT Name, Salary, DepartmentId,

									@rank := IF(@pre_d = DepartmentId, @rank + (@pre_s <> Salary), 1) AS rank,

									@pre_d := DepartmentId, @pre_s := Salary 

									FROM Employee, (SELECT @pre_d := -1, @pre_s := -1, @rank := 1) AS init

									ORDER BY DepartmentId, Salary DESC) e JOIN Department d ON e.DepartmentId = d.Id

									WHERE e.rank <= 3 ORDER BY d.Name, e.Salary DESC;