https://leetcode.com/problems/exchange-seats/
Mary is a teacher in a middle school and she has a table
The column id is continuous increment.seat
storing students' names and their corresponding seat ids.+---------+---------+ | 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.
If the number of students is odd, there is no need to change the last one's seat.
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;https://www.cnblogs.com/wxisme/p/7520231.html
SELECT s.id, s.student FROM ( SELECT id - 1 AS id, student FROM seat WHERE (id % 2 = 0) UNION SELECT (CASE WHEN (cnt%2=1) AND id=cnt THEN id ELSE id + 1 END) AS id, student FROM seat, (select count(*) as cnt from seat) as seatcnt WHERE (id % 2 = 1) ) s GROUP BY s.id ASC
select case when id = (select max(id) from seat) and mod(id, 2) = 1 then id when id < (select max(id) from seat) and mod(id, 2) = 1 then id + 1 when mod(id, 2) = 0 then id - 1 end as id, student from seat order by id