LeetCode（数据库）- 将工资相同的雇员分组

-- 解决方案(1)
select A.employee_id,
       A.name,
       A.salary,
       dense_rank() over(order by A.salary) team_id
  from (select a.*, COUNT(1) over(partition by a.salary) cn from Employees a) A
 WHERE A.cn > 1
 order by team_id,employee_id
-- 解决方案(2)
WITH t AS(SELECT 1 no, employee_id, name, salary
FROM Employees
WHERE employee_id NOT IN (SELECT employee_id FROM Employees GROUP BY salary HAVING COUNT(*) = 1))
SELECT employee_id, name, salary, DENSE_RANK() OVER(PARTITION BY no ORDER BY salary) team_id
FROM t