SQL实现LeetCode(177.第N高薪水)

[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：

				?

									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