[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.

這道題是之前那道Second Highest Salary的拓展，根據(jù)之前那道題的做法，我們可以很容易的將其推展為N，根據(jù)對Second Highest Salary中解法一的分析，我們只需要將OFFSET后面的1改為N-1就行了，但是這樣MySQL會報錯，估計不支持運算，那么我們可以在前面加一個SET N = N - 1，將N先變成N-1再做也是一樣的：

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

根據(jù)對Second Highest Salary中解法四的分析，我們只需要將其1改為N-1即可，這里卻支持N-1的計算，參見代碼如下：

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

當(dāng)然我們也可以通過將最后的>改為>=，這樣我們就可以將N-1換成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

類似題目：

Second Highest Salary

參考資料：

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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