leetcode185 部门工资前三高的所有员工 Department Top Three Salaries

时间：2019-10-22 23:51:05 阅读：96 评论：0 收藏：0 [点我收藏+]

Employee 表包含所有员工信息，每个员工有对应的 Id，此外还有一列部门 Id。

技术图片

创建表和数据：

Create table If Not Exists Employee (Idint, Name varchar(255), Salary int, DepartmentId int);
Create table If Not Exists Department (Idint, Name varchar(255));
Truncate table Employee;
insert into Employee (Id, Name, Salary,DepartmentId) values (‘1‘, ‘Joe‘, ‘70000‘, ‘1‘);
insert into Employee (Id, Name, Salary,DepartmentId) values (‘2‘, ‘Henry‘, ‘80000‘, ‘2‘);
insert into Employee (Id, Name, Salary,DepartmentId) values (‘3‘, ‘Sam‘, ‘60000‘, ‘2‘);
insert into Employee (Id, Name, Salary,DepartmentId) values (‘4‘, ‘Max‘, ‘90000‘, ‘1‘);
insert into Employee (Id, Name, Salary,DepartmentId) values (‘5‘, ‘Janet‘, ‘69000‘, ‘1‘);
insert into Employee (Id, Name, Salary,DepartmentId) values (‘6‘, ‘Randy‘, ‘85000‘, ‘1‘);
Truncate table Department;
insert into Department (Id, Name) values(‘1‘, ‘IT‘);
insert into Department (Id, Name) values(‘2‘, ‘Sales‘);

解法：

1.判断每个人A是不是在这三批人中的一个。找出同一部门种比A薪水高的薪水种数N。用子查询完成。如果N<3，那么A属于这三批人。

select D.name as Department,E.name as Employee,E.Salary
from Employee as E join Department as D on (E.departmentid = D.id)
where (
    select count(distinct E1.salary)
    from Employee as E1
    where E1.departmentid = E.departmentid and E1.salary > E.salary
) <3

2.先找出每个部门薪水第三高的薪水A。每个人的薪水只要大于等于A，他肯定在这三批人中

SELECT *
FROM Employee e1
LEFT JOIN Employee e2 ON(e1.DepartmentId=e2.DepartmentId AND e1.Salary>e2.Salary)
LEFT JOIN Employee e3 ON(e2.DepartmentId=e3.DepartmentId AND e2.Salary>e3.Salary);

+------+-------+--------+--------------+------+-------+--------+--------------+------+-------+--------+--------------+
| Id   | Name  | Salary | DepartmentId | Id   | Name  | Salary | DepartmentId | Id   | Name  | Salary | DepartmentId |
+------+-------+--------+--------------+------+-------+--------+--------------+------+-------+--------+--------------+
|    4 | Max   |  90000 |            1 |    1 | Joe   |  85000 |            1 |    5 | Janet |  69000 |            1 |
|    4 | Max   |  90000 |            1 |    6 | Randy |  85000 |            1 |    5 | Janet |  69000 |            1 |
|    1 | Joe   |  85000 |            1 |    7 | Will  |  70000 |            1 |    5 | Janet |  69000 |            1 |
|    4 | Max   |  90000 |            1 |    7 | Will  |  70000 |            1 |    5 | Janet |  69000 |            1 |
|    6 | Randy |  85000 |            1 |    7 | Will  |  70000 |            1 |    5 | Janet |  69000 |            1 |
|    4 | Max   |  90000 |            1 |    1 | Joe   |  85000 |            1 |    7 | Will  |  70000 |            1 |
|    4 | Max   |  90000 |            1 |    6 | Randy |  85000 |            1 |    7 | Will  |  70000 |            1 |
|    2 | Henry |  80000 |            2 |    3 | Sam   |  60000 |            2 | NULL | NULL  |   NULL |         NULL |
|    1 | Joe   |  85000 |            1 |    5 | Janet |  69000 |            1 | NULL | NULL  |   NULL |         NULL |
|    4 | Max   |  90000 |            1 |    5 | Janet |  69000 |            1 | NULL | NULL  |   NULL |         NULL |
|    6 | Randy |  85000 |            1 |    5 | Janet |  69000 |            1 | NULL | NULL  |   NULL |         NULL |
|    7 | Will  |  70000 |            1 |    5 | Janet |  69000 |            1 | NULL | NULL  |   NULL |         NULL |
|    3 | Sam   |  60000 |            2 | NULL | NULL  |   NULL |         NULL | NULL | NULL  |   NULL |         NULL |
|    5 | Janet |  69000 |            1 | NULL | NULL  |   NULL |         NULL | NULL | NULL  |   NULL |         NULL |
+------+-------+--------+--------------+------+-------+--------+--------------+------+-------+--------+--------------+

从结果中发现，求第三高的薪水，只能在e3.Salary上求max。且要处理 e3.Salary 为null的情况。如果在 e1.Salary 上求max，得到的一定是每个部门的最高薪水。因此left join 左边的表的所有元组必然在结果中。

用CASE WHEN END子句，对null字段进行处理。

在这里当值为NULL时，将其替换为0。

SELECT e1.DepartmentId, 
CASE 
    WHEN MAX(e3.salary) IS NULL THEN 0 
    ELSE MAX(e3.salary) 
END AS max_salary
FROM Employee e1
LEFT JOIN Employee e2 ON(e1.DepartmentId=e2.DepartmentId AND e1.Salary>e2.Salary)
LEFT JOIN Employee e3 ON(e2.DepartmentId=e3.DepartmentId AND e2.Salary>e3.Salary)
GROUP BY e1.DepartmentId

上面说顺序很重要，那么如果将上面的小于号的顺序，改为e1.Salary<e2.Salary<e3.Salary。再求e1.Salary的max，得出的值一定时每个部门最高的薪水，而不是第三高的薪水。因为left join最左边的表的元组一定全部都在！

将上面的结果与员工表和部门表，再连接起来，得出结果。

SELECT D.name AS Department,E.name AS Employee,E.Salary
FROM Employee AS E
JOIN Department AS D ON (E.departmentid = D.id)
JOIN (
    SELECT e1.DepartmentId, CASE WHEN MAX(e3.salary) IS NULL THEN 0 ELSE MAX(e3.salary) END AS m
    FROM Employee e1
    LEFT JOIN Employee e2 ON(e1.DepartmentId=e2.DepartmentId AND e1.Salary>e2.Salary)
    LEFT JOIN Employee e3 ON(e2.DepartmentId=e3.DepartmentId AND e2.Salary>e3.Salary)
    GROUP BY e1.DepartmentId
) AS F ON (E.departmentid = F.departmentid AND E.salary >= F.m)

3.延续解法二的思路。应用用户变量，求每个部门第三高薪水。

定义三个用户变量：@pre_salary——上一行的薪水；@pre_deptid——上一行的部门id；@salary_cnt——第几种薪水。

SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0;

其初始值如下，构成了一个表。

+--------------------+--------------------+----------------+
| @pre_salary:= NULL | @pre_deptid:= NULL | @salary_cnt:=0 |
+--------------------+--------------------+----------------+
| NULL               | NULL               |              0 |
+--------------------+--------------------+----------------+

将这样的表命名为A。

(
    SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
) 
AS A

先看将员工表与表A叉积。

select *
FROM Employee t, 
(
    SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
) 
AS A

结果为

+------+-------+--------+--------------+--------------------+--------------------+----------------+
| Id   | Name  | Salary | DepartmentId | @pre_salary:= NULL | @pre_deptid:= NULL | @salary_cnt:=0 |
+------+-------+--------+--------------+--------------------+--------------------+----------------+
|    1 | Joe   |  85000 |            1 | NULL               | NULL               |              0 |
|    2 | Henry |  80000 |            2 | NULL               | NULL               |              0 |
|    3 | Sam   |  60000 |            2 | NULL               | NULL               |              0 |
|    4 | Max   |  90000 |            1 | NULL               | NULL               |              0 |
|    5 | Janet |  69000 |            1 | NULL               | NULL               |              0 |
|    6 | Randy |  85000 |            1 | NULL               | NULL               |              0 |
|    7 | Will  |  70000 |            1 | NULL               | NULL               |              0 |
+------+-------+--------+--------------+--------------------+--------------------+----------------+

但这样的结果，对找每个部门的不同薪水没有帮助。要对结果按部门id递增排序，再按薪水降序。

select *
FROM Employee t, 
(
    SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
) 
AS A
ORDER BY t.DepartmentId, t.Salary DESC

排序后

+------+-------+--------+--------------+--------------------+--------------------+----------------+
| Id   | Name  | Salary | DepartmentId | @pre_salary:= NULL | @pre_deptid:= NULL | @salary_cnt:=0 |
+------+-------+--------+--------------+--------------------+--------------------+----------------+
|    4 | Max   |  90000 |            1 | NULL               | NULL               |              0 |
|    1 | Joe   |  85000 |            1 | NULL               | NULL               |              0 |
|    6 | Randy |  85000 |            1 | NULL               | NULL               |              0 |
|    7 | Will  |  70000 |            1 | NULL               | NULL               |              0 |
|    5 | Janet |  69000 |            1 | NULL               | NULL               |              0 |
|    2 | Henry |  80000 |            2 | NULL               | NULL               |              0 |
|    3 | Sam   |  60000 |            2 | NULL               | NULL               |              0 |
+------+-------+--------+--------------+--------------------+--------------------+----------------+

从这个结果中找出，每个部门的不同薪水个数。

对结果中的每行，执行下述逻辑，计数。

if (当前行的salary = @pre_salary and 当前行的departmentid = @pre_deptid)
{
    //相同的薪水
    @salary_cnt 不变
}
else if(当前行的departmentid = @pre_deptid)
{
    //不同的薪水
    @salary_cnt = @salary_cnt + 1
}
else
{
    //不同的部门，计数重新从1开始
    @salary_cnt = 1
}
//更新pre_salary和pre_deptid
@pre_salary = 当前行的salary
@pre_deptid = 当前行的departmentid

将此逻辑翻译为SQL代码，结果命令为表B

(
    SELECT
    @salary_cnt:= IF(t.Salary = @pre_salary AND t.DepartmentId = @pre_deptid, 
                @salary_cnt, 
            IF(t.DepartmentId = @pre_deptid, 
                             @salary_cnt + 1, 
                             1)
        ) 
        AS `cnt`
    ,@pre_salary := t.Salary AS `salary`
    ,@pre_deptid := t.DepartmentId AS `deptid`
    FROM Employee t, 
    (
        SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
    ) 
    AS A
    ORDER BY t.DepartmentId, t.Salary DESC
) 
AS B

结果为：

+------+--------+--------+
| cnt  | salary | deptid |
+------+--------+--------+
|    1 |  90000 |      1 |
|    2 |  85000 |      1 |
|    2 |  85000 |      1 |
|    3 |  70000 |      1 |
|    4 |  69000 |      1 |
|    1 |  80000 |      2 |
|    2 |  60000 |      2 |
+------+--------+--------+

从表B中，选出每个部门中前三种薪水，取最小的薪水。结果命名为C。

(
SELECT B.deptid,MIN(B.salary) AS `salary`
FROM 
    (
        SELECT
        @salary_cnt:= IF(t.Salary = @pre_salary AND t.DepartmentId = @pre_deptid, 
            @salary_cnt, 
                IF(t.DepartmentId = @pre_deptid, @salary_cnt + 1, 1)
            ) 
            AS `cnt`
        ,@pre_salary := t.Salary AS `salary`
        ,@pre_deptid := t.DepartmentId AS `deptid`
        FROM Employee t, 
        (
            SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
        ) 
        AS A
        ORDER BY t.DepartmentId, t.Salary DESC
    ) 
    AS B
WHERE B.cnt < 4
GROUP BY B.deptid
) AS C

结果为：

+--------+--------+
| deptid | salary |
+--------+--------+
|      1 |  70000 |
|      2 |  60000 |
+--------+--------+

再将表C与员工表和部门表连接，所有薪水大于等于C.salary都取出来。

SELECT D.name AS `Department`,E.name AS `Employee`,E.Salary
FROM Employee AS E
JOIN Department AS D ON (E.departmentid = D.id)
JOIN 
    (
    SELECT B.deptid,MIN(B.salary) AS `salary`
    FROM 
        (
            SELECT
            @salary_cnt:= IF(t.Salary = @pre_salary AND t.DepartmentId = @pre_deptid, 
                @salary_cnt, 
                    IF(t.DepartmentId = @pre_deptid, @salary_cnt + 1, 1)
                ) 
                AS `cnt`
            ,@pre_salary := t.Salary AS `salary`
            ,@pre_deptid := t.DepartmentId AS `deptid`
            FROM Employee t, 
            (
                SELECT @pre_salary:= NULL, @pre_deptid:= NULL, @salary_cnt:=0
            ) 
            AS A
            ORDER BY t.DepartmentId, t.Salary DESC
        ) 
        AS B
    WHERE B.cnt < 4
    GROUP BY B.deptid
    ) AS C
    ON (D.id = C.deptid AND E.salary >= C.salary)

4.计算每个员工的薪水，在其所在部门的薪水中的排名。最后取出每个部门中排名前三的员工。内在的逻辑与解法三殊途同归，但是具体的方法不同。

先取出每个部门不同薪水，并按部门id升序，薪水降序。结果命名为表A。

(
SELECT DepartmentId,Salary
FROM Employee
GROUP BY DepartmentId,Salary
ORDER BY DepartmentId,Salary DESC
)
AS A

结果为：

+--------------+--------+
| DepartmentId | Salary |
+--------------+--------+
|            1 |  90000 |
|            1 |  85000 |
|            1 |  70000 |
|            1 |  69000 |
|            2 |  80000 |
|            2 |  60000 |
+--------------+--------+

在表A上求出每个薪水的排名。也是借助于用户变量：@pre_deptid——上一行的部门id，@rank——此行的排名。

(SELECT @pre_deptid:=null, @rank:=0) AS B

叉积A和B，执行下面的逻辑计算@rank。

if (@pre_deptid = 当前行的departmentid)
{
    //同一部门
    @rank = @rank + 1
}
else
{
    //不同部门。@rank重置为1
    @rank=1
}

逻辑转换为：

CASE 
    WHEN @pre_deptid = DepartmentId THEN @rank:= @rank + 1
    WHEN @pre_deptid := DepartmentId THEN @rank:= 1
END AS `rank`

薪水排名逻辑，结果命名为

(
SELECT A.DepartmentId, A.Salary,
       CASE 
            WHEN @pre_deptid = DepartmentId THEN @rank:= @rank + 1
            WHEN @pre_deptid := DepartmentId THEN @rank:= 1
       END AS `rank`
FROM (SELECT @pre_deptid:=null, @rank:=0) 
       AS B,
     (
         SELECT DepartmentId,Salary
         FROM Employee
         GROUP BY DepartmentId,Salary
         ORDER BY DepartmentId,Salary DESC
     ) 
         AS A
)
AS C

结果为

+--------------+--------+------+
| DepartmentId | Salary | rank |
+--------------+--------+------+
|            1 |  90000 |    1 |
|            1 |  85000 |    2 |
|            1 |  70000 |    3 |
|            1 |  69000 |    4 |
|            2 |  80000 |    1 |
|            2 |  60000 |    2 |
+--------------+--------+------+

再将表C与员工表和部门表连接，取出排名小于等于3的员工。

SELECT D.Name as Department, E.NAME  AS Employee, E.Salary
FROM (
        SELECT A.DepartmentId, A.Salary,
               CASE 
                    WHEN @pre_deptid = DepartmentId THEN @rank:= @rank + 1
                    WHEN @pre_deptid := DepartmentId THEN @rank:= 1
               END AS `rank`
        FROM (SELECT @pre_deptid:=null, @rank:=0) 
               AS B,
             (
                 SELECT DepartmentId,Salary
                 FROM Employee
                 GROUP BY DepartmentId,Salary
                 ORDER BY DepartmentId,Salary DESC
             ) 
                 AS A
       )
       AS C
INNER JOIN Department AS D ON C.DepartmentId = D.Id
INNER JOIN Employee AS E ON C.DepartmentId = E.DepartmentId AND C.Salary = E.Salary AND C.rank <= 3

用CASE WHEN END子句，对null字段进行处理。

在这里当值为NULL时，将其替换为0。

leetcode185 部门工资前三高的所有员工 Department Top Three Salaries

标签：信息建表子查询递增 img 定时 not 排序翻译

原文地址：https://www.cnblogs.com/forever-fortunate/p/11723248.html

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