Solution


Approach I: Using flow control statement CASE [Accepted]

Algorithm

For students with odd id, the new id is (id+1) after switch unless it is the last seat. And for students with even id, the new id is (id-1). In order to know how many seats in total, we can use a subquery:

SELECT
    COUNT(*) AS counts
FROM
    seat

Then, we can use the CASE statement and MOD() function to alter the seat id of each student.

MySQL

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;

Approach II: Using bit manipulation and COALESCE() [Accepted]

Algorithm

Bit manipulation expression (id+1)^1-1 can calculate the new id after switch.

SELECT id, (id+1)^1-1, student FROM seat;
| id | (id+1)^1-1 | student |
|----|------------|---------|
| 1  | 2          | Abbot   |
| 2  | 1          | Doris   |
| 3  | 4          | Emerson |
| 4  | 3          | Green   |
| 5  | 6          | Jeames  |

Then, we can make a temp table and join seat with this table like below.

SELECT
    *
FROM
    seat s1
        LEFT JOIN
    seat s2 ON (s1.id+1)^1-1 = s2.id
ORDER BY s1.id;
| id | student | id | student |
|----|---------|----|---------|
| 1  | Abbot   | 2  | Doris   |
| 2  | Doris   | 1  | Abbot   |
| 3  | Emerson | 4  | Green   |
| 4  | Green   | 3  | Emerson |
| 5  | Jeames  |    |         |

Note:The first two columns are from s1 and the last two are from s2.

At last, we can output s1.id and s2.student. However, the s2.student is NULL for seat id '5' but s1.student is right. Thus, we we can use function COALESCE() to generate the correct output for the last record.

MySQL

SELECT
    s1.id, COALESCE(s2.student, s1.student) AS student
FROM
    seat s1
        LEFT JOIN
    seat s2 ON ((s1.id + 1) ^ 1) - 1 = s2.id
ORDER BY s1.id;

Note: This solution comes from @FANGXIAOFANG.