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

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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