[SQL]LeetCode626. 换座位 | Exchange Seats

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SQL架构

Create table If Not Exists seat(id int, student varchar(255))
Truncate table seat
insert into seat (id, student) values ('1', 'Abbot')
insert into seat (id, student) values ('2', 'Doris')
insert into seat (id, student) values ('3', 'Emerson')
insert into seat (id, student) values ('4', 'Green')
insert into seat (id, student) values ('5', 'Jeames')

---

Mary is a teacher in a middle school and she has a table seat storing students' names and their corresponding seat ids.

The column id is continuous increment. 

Mary wants to change seats for the adjacent students. 

Can you write a SQL query to output the result for Mary? 

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Abbot   |
|    2    | Doris   |
|    3    | Emerson |
|    4    | Green   |
|    5    | Jeames  |
+---------+---------+

For the sample input, the output is: 

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Doris   |
|    2    | Abbot   |
|    3    | Green   |
|    4    | Emerson |
|    5    | Jeames  |
+---------+---------+

Note:
If the number of students is odd, there is no need to change the last one's seat.

---

小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。

其中纵列的 id 是连续递增的

小美想改变相邻俩学生的座位。

你能不能帮她写一个 SQL query 来输出小美想要的结果呢？ 

示例：

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Abbot   |
|    2    | Doris   |
|    3    | Emerson |
|    4    | Green   |
|    5    | Jeames  |
+---------+---------+

假如数据输入的是上表，则输出结果如下：

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Doris   |
|    2    | Abbot   |
|    3    | Green   |
|    4    | Emerson |
|    5    | Jeames  |
+---------+---------+

注意：

如果学生人数是奇数，则不需要改变最后一个同学的座位。

---

Runtime: 152 ms

# Write your MySQL query statement below
select 
    (case when id % 2 = 0 then id - 1
        when id % 2 = 1 and id <> c.cnt then id + 1
        else id
     end) as id, student
from 
    seat,
    (select count(id) as cnt from seat) as c
order by id

---

253ms

# Write your MySQL query statement below
SELECT 
    (CASE
        WHEN MOD(id, 2) != 0 AND counts != id THEN id + 1
        WHEN MOD(id, 2) != 0 AND counts = id THEN id
        ELSE id - 1
    END) as id,
    student
FROM 
    seat,
    (SELECT COUNT(*) as counts
    FROM seat) as seat_counts
ORDER BY id ASC;

---

255ms

# Write your MySQL query statement below
select * from (
select case when b.Id is null then a.Id else b.Id end as Id,a.student from seat a left join seat b on a.Id+1=b.Id where mod(a.id,2)=1
union 
select b.Id,a.student from seat a left join seat b on a.Id-1=b.Id where mod(a.id,2)=0
) a order by a.id

---

257ms

select
    case
        when id%2=1 and id=(select max(id) from seat) then id
        when id%2=1 then id+1
        else id-1 end as id, 
    student
FROM seat
order by id

---

258ms

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

---

260ms

SELECT
    (CASE WHEN id%2=0 THEN id-1 
     WHEN id%2=1 AND id<(select max(id) FROM seat) THEN id+1 
    ELSE id
    END) as id
    ,student
FROM seat
ORDER BY id;