【LeetCode】185. 部門工資前三高的員工 學習筆記
阿新 • • 發佈：2019-01-25
Employee 表包含所有員工資訊，每個員工有其對應的 Id, salary 和 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 | +----+-------+--------+--------------+
Department 表包含公司所有部門的資訊。
+----+----------+ | Id | Name | +----+----------+ | 1 | IT | | 2 | Sales | +----+----------+
編寫一個 SQL 查詢，找出每個部門工資前三高的員工。例如，根據上述給定的表格，查詢結果應返回：
+------------+----------+--------+ | Department | Employee | Salary | +------------+----------+--------+ | IT | Max | 90000 | | IT | Randy | 85000 | | IT | Joe | 70000 | | Sales | Henry | 80000 | | Sales | Sam | 60000 | +------------+----------+--------+
例1：
- # Write your MySQL query statement below
- Select d.Nameas Department, e.Nameas Employee, e.Salary
- from Department d, Employee e
- where b.DepartmentId = d.Id and (
- Selectcount(distinct Salary) From Employee where DepartmentId=d.Id and Salary > e.Salary
- )<3
- orderby Department
例2：
- # Write your MySQL query statement below
- SELECT d.NameAS Department, e1.NameAS Employee, e1.Salary
- FROM Employee e1 JOIN Department d ON e1.DepartmentId = d.Id
- WHERE 3 > (SELECTCOUNT(DISTINCT e2.Salary) FROM Employee e2
- WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId);
以例2：基本思路是通過兩個employee的表，進行對比，約束條件是 departmentid 相等and salary 最大的。
而小於三則是對數量的約束。
例3：
- # Write your MySQL query statement below
- SELECT D1.Name Department, E1.Name Employee, E1.Salary
- FROM Employee E1, Employee E2, Department D1
- WHERE E1.DepartmentID = E2.DepartmentID
- AND E2.Salary >= E1.Salary
- AND E1.DepartmentID = D1.ID
- GROUPBY E1.Name
- HAVINGCOUNT(DISTINCT E2.Salary) <= 3
- ORDERBY D1.Name, E1.Salary DESC;