176. Second Highest Salary 第二高的薪水

题目链接
tag:

  • Easy;

question
  SQL Schema

Create table If Not Exists Employee (Id int, Salary int)
Truncate table Employee
insert into Employee (Id, Salary) values ('1', '100')
insert into Employee (Id, Salary) values ('2', '200')
insert into Employee (Id, Salary) values ('3', '300')

Write a SQL query to get the second highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+

For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+

思路:
  这道题让我们找表中某列第二大的数,这道题有很多种解法,先来看一种使用Limit和Offset两个关键字的解法,MySQL中Limit后面的数字限制了我们返回数据的个数,Offset是偏移量,那么如果我们想找第二高薪水,我们首先可以先对薪水进行降序排列,然后我们将Offset设为1,那么就是从第二个开始,也就是第二高薪水,然后我们将Limit设为1,就是只取出第二高薪水,如果将Limit设为2,那么就将第二高和第三高薪水都取出来:

解法一:

# Write your MySQL query statement below
SELECT Salary FROM Employee GROUP BY Salary
UNION ALL (SELECT NULL AS Salary)
ORDER BY Salary DESC LIMIT 1 OFFSET 1;

我们也可以使用Max函数来做,这个返回最大值,逻辑是我们取出的不包含最大值的数字中的最大值,即为第二大值:

解法二:

SELECT MAX(Salary) FROM Employee 
WHERE Salary NOT IN
(SELECT MAX(Salary) FROM Employee);

下面这种方法和上面基本一样,就是用小于号<代替了Not in关键字,效果相同:

解法三:

SELECT MAX(Salary) FROM Employee
Where Salary <
(SELECT MAX(Salary) FROM Employee);

  最后来看一种可以扩展到找到第N高的薪水的方法,只要将下面语句中的1改为N-1即可,第二高的薪水带入N-1就是1,下面语句的逻辑是,假如我们要找第二高的薪水,那么我们允许其中一个最大值存在,然后在其余的数字中找出最大的,即为整个的第二大的值;

解法四:

SELECT MAX(Salary) FROM Employee E1
WHERE 1 =
(SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary > E1.Salary);

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