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：
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
本文转自博客园Grandyang的博客，原文链接：第N高薪水[LeetCode] Nth Highest Salary ，如需转载请自行联系原博主。
