LeetCode（数据库）- 员工薪水中位数

题目链接：​​点击打开链接​​

题目大意：略。

解题思路：解决方案(2)：对于一个奇数 / 偶数长度数组中的中位数，大于这个数的数值个数等于小于这个数的数值个数。最后需要“GROUP BY Company, Salary”是因为子连接时会有左右重复的情况。

AC 代码

-- 解决方案(1)
SELECT id, company, salary
FROM
(
    SELECT id, company, salary,
    ROW_NUMBER() OVER (PARTITION BY company ORDER BY Salary ASC, id ASC) AS row_num,
    COUNT(Id) OVER (PARTITION BY company) AS count_id
    FROM Employee
)
WHERE row_num IN (FLOOR((count_id + 1)/2), FLOOR((count_id + 2)/2));

-- 解决方案(2)
SELECT Id, Company, Salary
FROM Employee
WHERE Id in (
    SELECT e1.Id
    FROM Employee e1
    JOIN Employee e2
    ON e1.Company = e2.Company
    GROUP BY e1.Id
    HAVING SUM(CASE WHEN e1.Salary >= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2 
    AND SUM(CASE WHEN e1.Salary <= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2
    )
GROUP BY Company, Salary
ORDER BY Company;