176. Second Highest Salary(Leetcode)
文章出處:http://www.cnblogs.com/grandyang/p/5348961.html
Write a SQL query to get the second highest salary from the Employee
table.
+----+--------+ | Id | Salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+
For example, given the above Employee table, the second highest salary is 200
null
.
這道題讓我們找表中某列第二大的數,這道題有很多種解法,先來看一種使用Limit和Offset兩個關鍵字的解法,MySQL中Limit後面的數字限制了我們返回資料的個數,Offset是偏移量,那麼如果我們想找第二高薪水,我們首先可以先對薪水進行降序排列,然後我們將Offset設為1,那麼就是從第二個開始,也就是第二高薪水,然後我們將Limit設為1,就是隻取出第二高薪水,如果將Limit設為2,那麼就將第二高和第三高薪水都取出來:
解法一:
SELECT Salary as SecondHighestSalary FROM Employee GROUP BY Salary UNION ALL (SELECT NULL AS Salary) ORDER BY Salary DESC LIMIT 1 OFFSET 1;
我們也可以使用Max函式來做,這個返回最大值,邏輯是我們取出的不包含最大值的數字中的最大值,即為第二大值:
解法二:
SELECT MAX(Salary) FROM Employee WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee);
下面這種方法和上面基本一樣,就是用小於號<代替了Not in關鍵字,效果相同:
解法三:
SELECT MAX(Salary) FROM Employee Where Salary < (SELECT MAX(Salary) FROM Employee);
最後來看一種可以擴充套件到找到第N高的薪水的方法,只要將下面語句中的1改為N-1即可,第二高的薪水帶入N-1就是1,下面語句的邏輯是,假如我們要找第二高的薪水,那麼我們允許其中一個最大值存在,然後在其餘的數字中找出最大的,即為整個的第二大的值;
Distinct是去重操作,一般放在開頭(緊跟select後)
解法四理解:要找到第二高的薪水,在表中就存在一個1個比它大的值,E1.Salary < E2.Salary 縮小了查詢範圍,將E1表第一高的薪水捨去,因為在表中沒有比他小的值。其實連MAX都不用加,因為最後得出的表中只有一個值(除非有重複值),但是leetcode要求不存在第二打返回null,只有聚合函式能返回null值,如果不加MAX,Salary不存在會返回空白(leetcode判定)
解法四:
SELECT MAX(Salary) FROM Employee E1 WHERE 1 = (SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary > E1.Salary);
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