[LeetCode] 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：

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

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