跳转至

176. 第二高的薪水

题目描述

Employee 表:

+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| salary      | int  |
+-------------+------+
在 SQL 中,id 是这个表的主键。
表的每一行包含员工的工资信息。

 

查询并返回 Employee 表中第二高的薪水 。如果不存在第二高的薪水,查询应该返回 null(Pandas 则返回 None)

查询结果如下例所示。

 

示例 1:

输入:
Employee 表:
+----+--------+
| id | salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+
输出:
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

示例 2:

输入:
Employee 表:
+----+--------+
| id | salary |
+----+--------+
| 1  | 100    |
+----+--------+
输出:
+---------------------+
| SecondHighestSalary |
+---------------------+
| null                |
+---------------------+

解法

方法一:使用 LIMIT 语句和子查询

我们可以按照薪水降序排列,然后使用 LIMIT 语句来获取第二高的薪水,如果不存在第二高的薪水,那么就返回 null

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
import pandas as pd


def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
    # Drop any duplicate salary values to avoid counting duplicates as separate salary ranks
    unique_salaries = employee["salary"].drop_duplicates()

    # Sort the unique salaries in descending order and get the second highest salary
    second_highest = (
        unique_salaries.nlargest(2).iloc[-1] if len(unique_salaries) >= 2 else None
    )

    # If the second highest salary doesn't exist (e.g., there are fewer than two unique salaries), return None
    if second_highest is None:
        return pd.DataFrame({"SecondHighestSalary": [None]})

    # Create a DataFrame with the second highest salary
    result_df = pd.DataFrame({"SecondHighestSalary": [second_highest]})

    return result_df
1
2
3
4
5
6
7
8
# Write your MySQL query statement below
SELECT
    (
        SELECT DISTINCT salary
        FROM Employee
        ORDER BY salary DESC
        LIMIT 1, 1
    ) AS SecondHighestSalary;

方法二:使用 MAX() 函数和子查询

我们也可以使用 MAX() 函数,从小于 MAX() 的薪水中挑选一个最大的薪水即可。

1
2
3
4
# Write your MySQL query statement below
SELECT MAX(salary) AS SecondHighestSalary
FROM Employee
WHERE salary < (SELECT MAX(salary) FROM Employee);

方法三:使用 DISTINCT 和窗口函数

我们还可以先通过 DENSE_RANK() 函数计算出每个员工的薪水排名,然后再筛选出排名为 $2$ 的员工薪水即可。

1
2
3
# Write your MySQL query statement below
WITH T AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rk FROM Employee)
SELECT (SELECT DISTINCT salary FROM T WHERE rk = 2) AS SecondHighestSalary;

评论