SQL实现LeetCode(177.第N高薪水)

时间:2022-11-04 15:16:25

[LeetCode] 177.Nth Highest Salary 第N高薪水

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

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

For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.

这道题是之前那道Second Highest Salary的拓展,根据之前那道题的做法,我们可以很容易的将其推展为N,根据对Second Highest Salary中解法一的分析,我们只需要将OFFSET后面的1改为N-1就行了,但是这样MySQL会报错,估计不支持运算,那么我们可以在前面加一个SET N = N - 1,将N先变成N-1再做也是一样的:

解法一:

?
1
2
3
4
5
6
7
8
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  SET N = N - 1;
  RETURN (
      SELECT DISTINCT Salary FROM Employee GROUP BY Salary
      ORDER BY Salary DESC LIMIT 1 OFFSET N
  );
END

根据对Second Highest Salary中解法四的分析,我们只需要将其1改为N-1即可,这里却支持N-1的计算,参见代码如下:

解法二:

?
1
2
3
4
5
6
7
8
9
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  RETURN (
      SELECT MAX(Salary) FROM Employee E1
      WHERE N - 1 =
      (SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
      WHERE E2.Salary > E1.Salary)
  );
END

当然我们也可以通过将最后的>改为>=,这样我们就可以将N-1换成N了:

解法三:

?
1
2
3
4
5
6
7
8
9
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  RETURN (
      SELECT MAX(Salary) FROM Employee E1
      WHERE N =
      (SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
      WHERE E2.Salary >= E1.Salary)
  );
END

类似题目:

Second Highest Salary

参考资料:

https://leetcode.com/discuss/88875/simple-answer-with-limit-and-offset

https://leetcode.com/discuss/63183/fastest-solution-without-using-order-declaring-variables

到此这篇关于SQL实现LeetCode(177.第N高薪水)的文章就介绍到这了,更多相关SQL实现第N高薪水内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家!

原文链接:https://www.cnblogs.com/grandyang/p/5348976.html