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:
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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