8000 solve problem Exchange Seats · P-ppc/leetcode@e1e4783 · GitHub
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solve problem Exchange Seats
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‎README.md

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|183|[Customers Who Never Order](https://leetcode-cn.com/problems/customers-who-never-order/description/)|[mysql](./database/CustomersWhoNeverOrder)|Easy|
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|197|[Rising Temperature](https://leetcode-cn.com/problems/rising-temperature/description/)|[mysql](./database/RisingTemperature)|Easy|
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|181|[Employees Earning More Than Their Managers](https://leetcode-cn.com/problems/employees-earning-more-than-their-managers/description/)|[mysql](./database/EmployeesEarningMoreThanTheirManagers)|Easy|
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|176|[Second Highest Salary](https://leetcode-cn.com/problems/second-highest-salary/description/)|[mysql](./database/SecondHighestSalary)|Easy|
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|176|[Second Highest Salary](https://leetcode-cn.com/problems/second-highest-salary/description/)|[mysql](./database/SecondHighestSalary)|Easy|
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|626|[Exchange Seats](https://leetcode-cn.com/problems/exchange-seats/description/)|[mysql](./database/ExchangeSeats)|Medium|

‎database/ExchangeSeats/README.md

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# Exchange Seats
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We can solve problem this problem by CASE function, or use union to solve it like below:
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```sql
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# Write your MySQL query statement below
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select id, student from (
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select id - 1 as id, student from seat where mod(id, 2) = 0 union
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select id + 1 as id, student from seat where mod(id, 2) = 1 and id != (select count(*) from seat) union
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select id, student from seat where mod(id, 2) = 1 and id = (select count(*) from seat)
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) as s order by id
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```

‎database/ExchangeSeats/solution.sql

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# Write your MySQL query statement below
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select (
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case
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when mod(id, 2) = 1 and id = counts then id
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when mod(id, 2) = 1 and id < counts then id + 1
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else id - 1
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end) as id, student from seat, (select count(*) as counts from seat) as seat_counts order by id

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