MySQL之LeetCode-184(部门工资最高的员工)

题目（中等）

Employee 表包含所有员工信息，每个员工有其对应的 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 |
±—±------±-------±-------------+
Department 表包含公司所有部门的信息。

±—±---------+
| Id | Name |
±—±---------+
| 1 | IT |
| 2 | Sales |
±—±---------+
编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

±-----------±---------±-------+
| Department | Employee | Salary |
±-----------±---------±-------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
±-----------±---------±-------+

解题语句

SELECT
  t.Name AS Department,
  e.Name AS Employee,
  e.Salary
FROM
  Employee AS e
  JOIN
    (SELECT
      MAX(e.Salary) AS Salary,
      d.Name,
      d.Id
    FROM
      Department AS d
      LEFT JOIN Employee AS e
        ON d.Id = e.DepartmentId
    GROUP BY e.DepartmentId) AS t
    ON e.Salary = t.Salary
    AND e.DepartmentId = t.Id

这个解法就是先查出每个部门的最高工资，然后再与员工表关联，找到员工名称。
我认为这种解法并不好，但是看了评论也没有看到怎么简洁的解法。
评论中的其它解法：

select
  d.Name as Department,
  e.Name as Employee,
  e.Salary
from
  Department d
  inner join Employee e
    on d.Id = e.DepartmentId
    and e.Salary >=
    (select
      max(Salary)
    from
      Employee
    where DepartmentId = d.Id)

最后的大于等于可以直接使用等于。因为后面已经是max了，也就不会大于了。

SELECT
  d.name AS Department,
  a.name AS Employee,
  a.Salary
FROM
  (SELECT
    e.*
  FROM
    employee e
    LEFT JOIN
      (SELECT
        MAX(e1.salary) AS salary,
        e1.DepartmentId
      FROM
        employee e1
      GROUP BY e1.DepartmentId) e2
      ON e.salary = e2.salary
      AND e.departmentid = e2.departmentid
  WHERE e2.salary IS NOT NULL) a
  LEFT JOIN department d
    ON a.departmentid = d.id
WHERE d.id IS NOT NULL

上面这个解法我测试时时间最短

SELECT
  d.name AS department,
  e.name AS employee,
  e.salary
FROM
  employee e,
  department d
WHERE e.departmentid = d.id
  AND (e.salary, e.departmentid) IN
  (SELECT
    MAX(e.salary),
    e.departmentid
  FROM
    employee e
  GROUP BY departmentid);

这个解法思路很奇特