177. Nth Highest Salary

置顶

Created At : 2023-10-28 15:14

Count:393 Views 👀 :

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

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

For example, given the above Employee table, the n th highest salary where n = 2 is 200. If there is no n th 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：

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：

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：

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

LeetCode All in One 题目讲解汇总(持续更新中…)

转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com