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;
类似题目：
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
LeetCode All in One 题目讲解汇总(持续更新中…)
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