LeetCode MySQL 569. 员工薪水中位数（over窗口函数）

1. 题目

Employee 表包含所有员工。Employee 表有三列：员工Id，公司名和薪水。

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|1    | A          | 2341   |
|2    | A          | 341    |
|3    | A          | 15     |
|4    | A          | 15314  |
|5    | A          | 451    |
|6    | A          | 513    |
|7    | B          | 15     |
|8    | B          | 13     |
|9    | B          | 1154   |
|10   | B          | 1345   |
|11   | B          | 1221   |
|12   | B          | 234    |
|13   | C          | 2345   |
|14   | C          | 2645   |
|15   | C          | 2645   |
|16   | C          | 2652   |
|17   | C          | 65     |
+-----+------------+--------+

请编写SQL查询来查找每个公司的薪水中位数。

挑战点：你是否可以在不使用任何内置的SQL函数的情况下解决此问题。

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|5    | A          | 451    |
|6    | A          | 513    |
|12   | B          | 234    |
|9    | B          | 1154   |
|14   | C          | 2645   |
+-----+------------+--------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/median-employee-salary
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 算出每个公司的人数、薪水的排序
• where 选择，根据人数奇偶筛选

# Write your MySQL query statement below

select Id, Company, Salary
from
(
    select *, 
            row_number() over (partition by Company order by Salary) rnk,
            count(*) over (partition by Company) num
    from Employee
) t
where(
        (num%2=1 and rnk = floor(num/2)+1)
        or
        (num%2=0 and (rnk = floor(num/2) or rnk = floor(num/2)+1))
     )

---

我的CSDN博客地址 https://michael.blog.csdn.net/

长按或扫码关注我的公众号（Michael阿明），一起加油、一起学习进步！