SQL实现LeetCode(176.第二高薪水)

[LeetCode] 176.Second Highest Salary 第二高薪水

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 second highest salary is 200. If there is no second highest salary, then the query should return null.

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:

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);

参考资料:

https://leetcode.com/discuss/47041/very-very-simple-solution

https://leetcode.com/discuss/42849/general-solution-not-using-max

https://leetcode.com/discuss/21751/simple-query-which-handles-the-null-situation

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