569. 员工薪水中位数

最新推荐文章于 2022-04-02 09:01:00 发布

偲偲粑

最新推荐文章于 2022-04-02 09:01:00 发布

阅读量385

点赞数

CC 4.0 BY-SA版权

分类专栏： MYSQL数据库LeetCode 刷题

本文链接：https://blog.youkuaiyun.com/weixin_43329319/article/details/95813708

MYSQL数据库LeetCode 刷题专栏收录该内容

58 篇文章

订阅专栏

该博客围绕力扣（LeetCode）的题目展开，介绍了解题思路。包括对公司收入进行正序和倒序排名，取倒序排名前两名的正序排名处理成中位数位置并去重，还进行了图文详解，最后给出了提交记录，用时352ms。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1、题目描述

在这里插入图片描述

来源：力扣（LeetCode）

2、解题思路

这道题真是让我很费神，主要是不喜欢那么多的嵌套

一、文字说明

1# 对每个公司的收入增加正序排名_rank(作为定位）和倒序排名rank_desc
2# 然后取倒序排名rank_desc<=2即前两名，的正序排名_rank处理成为，该公司的中位数位置，并去重:select distinct Company,case when _rank%2=0 then _rank/2+1 else _rank/2+0.5 end as mean，作为表add_mean
3# 另进行一次正序排名_rank，作为表add_rank
4# 2表联查
注意1：正序排名后的倒序排名操作，不能order by Salary，必须order by _rank，因为如果Salary相同，正倒序结果不符合预期，
注意2：自定义变量众多，必须各不相同；另rank为关键词

二、图文详解

正序排名，参考185. 部门工资前三高的员工
源数据：在这里插入图片描述
以下代码进行正序排名，结果见图：

select e.*,@i:=case when @j=@j:=Company then @i+1 else 1 end as _rank from Employee e,
(select @i:=1,@j:=-1) a order by Company,Salary

在这里插入图片描述
然后增加嵌套进行降序排名，结果见图：
select add_rank.*,@m:=case when @n=@n:=Company then @m+1 else 1 end as rank_desc from(
select e.,@i:=case when @j=@j:=Company then @i+1 else 1 end as _rank from Employee e,(select @i:=1,@j:=-1) a order by Company,Salary
)add_rank,(select @m:=1,@n:=-1) a order by Company,_rank desc

取rank_desc即排位的前两位的_rank值（可视为公司总个数），如果整除2则除以2+1，否则除以2+0.5，然后去重（此处不展开），结果见图：
select distinct Company,case when _rank%2=0 then _rank/2+1 else _rank/2+0.5 end as mean from (
select add_rank.,@m:=case when @n=@n:=Company then @m+1 else 1 end as rank_desc from
(select e.*,@i:=case when @j=@j:=Company then @i+1 else 1 end as _rank from Employee e,
(select @i:=1,@j:=-1) a order by Company,Salary) add_rank,(select @m:=1,@n:=-1) a order by Company,_rank desc
) add_rank_desc where rank_desc<=2
在这里插入图片描述

3、提交记录

我这样写法应该还清楚^_^,后面还附有刷完格式的一份

select add_rank.Id,add_rank.Company,add_rank.Salary
from
    (select distinct Company,case when _rank%2=0 then _rank/2+1 else _rank/2+0.5  end as mean
    from
        (select add_rank.*,@m:=case when @n=@n:=Company then @m+1 else 1 end as rank_desc
        from
            (select e.*,@i:=case when @j=@j:=Company then @i+1 else 1 end as _rank
             from Employee e,(select @i:=1,@j:=-1) a order by Company,Salary) add_rank,
            (select @m:=1,@n:=-1) a order by Company,_rank desc)
        add_rank_desc
    where rank_desc<=2) add_mean,

    (select e.*,@k:=case when @l=@l:=Company then @k+1 else 1 end as _rank
    from Employee e,(select @k:=1,@l:=-1) a
    order by Company,Salary) add_rank
where add_mean.Company=add_rank.Company and add_mean.mean=add_rank._rank
order by Company,Salary

特意刷了一下格式

SELECT 
    add_rank.Id, add_rank.Company, add_rank.Salary
FROM
    (SELECT DISTINCT
        Company,
            CASE
                WHEN _rank % 2 = 0 THEN _rank / 2 + 1
                ELSE _rank / 2 + 0.5
            END AS mean
    FROM
        (SELECT 
        add_rank.*,
            @m:=CASE
                WHEN @n = @n:=Company THEN @m + 1
                ELSE 1
            END AS rank_desc
    FROM
        (SELECT 
        e.*,
            @i:=CASE
                WHEN @j = @j:=Company THEN @i + 1
                ELSE 1
            END AS _rank
    FROM
        Employee e, (SELECT @i:=1, @j:=- 1) a
    ORDER BY Company , Salary) add_rank, (SELECT @m:=1, @n:=- 1) a
    ORDER BY Company , _rank DESC) add_rank_desc
    WHERE
        rank_desc <= 2) add_mean,
    (SELECT 
        e.*,
            @k:=CASE
                WHEN @l = @l:=Company THEN @k + 1
                ELSE 1
            END AS _rank
    FROM
        Employee e, (SELECT @k:=1, @l:=- 1) a
    ORDER BY Company , Salary) add_rank
WHERE
    add_mean.Company = add_rank.Company
        AND add_mean.mean = add_rank._rank
ORDER BY Company , Salary

352ms
在这里插入图片描述