SQL实现LeetCode(185.系里前三高薪水) SQL实现LeetCode(185.系里前三高薪

[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 eJOIN Department d on e.DepartmentId = d.IdWHERE (SELECT COUNT(DISTINCT Salary) FROM Employee WHERE Salary > e.SalaryAND 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 dWHERE (SELECT COUNT(DISTINCT Salary) FROM Employee WHERE Salary > e.SalaryAND 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;

【SQL实现LeetCode(185.系里前三高薪水)】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 initORDER BY DepartmentId, Salary DESC) e JOIN Department d ON e.DepartmentId = d.IdWHERE e.rank <= 3 ORDER BY d.Name, e.Salary DESC;

类似题目：
Department Highest Salary
Second Highest Salary
Combine Two Tables
参考资料：
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
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