Second Highest Salary --leetCode

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 query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

https://leetcode.com/problems/second-highest-salary/description/

solution:

本想找第二高的就是子查詢找到 兩個最高的 而後取最小的那個
select Salary as SecondHighestSalary from
(select * from Employee order by Salary desc limit 2) temp
order by Salary  asc limit 1
可是執行 告訴我結果是錯的  原來題目的要求中沒有第二高的就返回 null

看來轉換思路很重要呀，當時想按照上面的方式 怎麼處理子查詢只有一條的狀況呀 真是陷入了本身的坑
而後按照取到max最大的，從比最大的小中的最大的，若是我要取第三大的 是否是比較坑
SELECT MAX(Salary) as SecondHighestSalary
FROM Employee
WHERE Salary < (SELECT MAX(Salary) FROM Employee);


而後去看看網上的大神都是怎麼處理的 這邊UNION ALL 和 UNION是同樣的吧
SELECT Salary as SecondHighestSalary FROM Employee GROUP BY Salary
UNION ALL (SELECT NULL AS SecondHighestSalary)
ORDER BY SecondHighestSalary DESC LIMIT 1,1;
若是是我 mysql估計就會卡在這邊怎麼處理null的問題出不來了
SELECT Salary as SecondHighestSalary FROM Employee GROUP BY Salary  DESC LIMIT 1,1

下面這個max的用法 和上面的殊途同歸
SELECT MAX(Salary) as SecondHighestSalary  FROM Employee
WHERE Salary NOT IN
(SELECT MAX(Salary) FROM Employee);

後面這我的寫的真直接呀
SELECT MAX(Salary) as SecondHighestSalary FROM Employee E1
WHERE 1 =
(SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary > E1.Salary);

git地址：https://github.com/woshiyexinjie/leetcode-xin