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;
您可能感兴趣的文章
- 05-31MySQL中的 inner join 和 left join的区别解析(小结果集驱动大结果集)
- 05-31MySQL索引失效十种场景与优化方案
- 05-31MYSQL 高级文本查询之regexp_like和REGEXP详解
- 05-31MySQL获取binlog的开始时间和结束时间(最新方法)
- 05-31MySQL索引查询的具体使用
- 05-31基于MySQL和Redis扣减库存的实践
- 05-31关于MySQL的存储过程与存储函数
- 05-31MySQL实战文章(非常全的基础入门类教程)
- 05-31MySQL Flink Watermark实现事件时间处理的关键技术
- 05-31MySQL Flink实时流处理的核心技术之窗口机制


阅读排行
推荐教程
- 05-30Navicat for MySQL 11注册码激活码汇总
- 05-27Mysql误删数据快速恢复
- 05-31VS2022连接数据库MySQL并进行基本的表的操作指南
- 05-30解决seata不能使用mysql8版本的问题方法
- 05-30MYSQL字符集设置的方法详解(终端的字符集)
- 05-30解决MySQL启动报错:ERROR 2003 (HY000): Can't con
- 05-30关于Mysql-connector-java驱动版本问题总结
- 11-22mac下安装mysql忘记密码的修改方法
- 05-30MySQL中的隐藏列的具体查看
- 11-22mysql exists与not exists实例详解