[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  |
+------------+----------+--------+

這道題是之前那道Department Highest Salary的拓展，難度標記為Hard，還是蠻有難度的一道題，綜合了前面很多題的知識點，首先看使用Select Count(Distinct)的方法，我們內交Employee和Department兩張表，然后我們找出比當前薪水高的最多只能有兩個，那么前三高的都能被取出來了，參見代碼如下：

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;

下面這種方法將上面方法中的3換成了IN (0, 1, 2)，是一樣的效果：

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;

或者我們也可以使用Group by Having Count(Distinct ..) 關鍵字來做：

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;

下面這種方法略微復雜一些，用到了變量，跟Consecutive Numbers中的解法三使用的方法一樣，目的是為了給每個人都按照薪水的高低增加一個rank，最后返回rank值小于等于3的項即可，參見代碼如下：

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;

類似題目：

Department Highest Salary

Second Highest Salary

Combine Two Tables

參考資料：

https://leetcode.com/discuss/23002/my-tidy-solution

https://leetcode.com/discuss/91087/yet-another-solution-using-having-count-distinct

https://leetcode.com/discuss/69880/two-solutions-1-count-join-2-three-variables-join

到此這篇關于SQL實現LeetCode(185.系里前三高薪水)的文章就介紹到這了,更多相關SQL實現系里前三高薪水內容請搜索腳本之家以前的文章或繼續瀏覽下面的相關文章希望大家以后多多支持腳本之家！