626. Exchange Seats-(LeetCode之Database篇)
问题表述
数据库表如下:
id | student |
---|---|
1 | Abbot |
2 | Doris |
3 | Emerson |
4 | Green |
5 | Jeames |
现在要通过SQL语句将表变换成如下:
id | student |
---|---|
1 | Doris |
2 | Abbot |
3 | Green |
4 | Emerson |
5 | Jeames |
即id不变,奇数位和偶数位交换位置,如果表的总行数为奇数,则最后一行不变。
问题解决
首先看到这个问题,我就想SQL里面是不是有什么置换函数之类的,结果去查了查,并没有这样的函数。在我尝试了各种select方法后,还是没能将这题解出来…最后还是去讨论区看了看大神们的解答,看完各种答案后瞬间豁然开朗。
这题的解题思路其实并不是想办法将student的列置换,而是通过操作id列来达到置换的效果。
比较通过的解法就是下面这种:
select
if(id < (select count(*) from seat), if(id mod 2=0, id-1, id+1), if(id mod 2=0, id-1, id)) as id, student
from seat
order by id asc
其中IF函数的用法:
格式:IF(Condition,A,B)
意义:当Condition为TRUE时,返回A;当Condition为FALSE时,返回B。
所以上面的SQL语句就是,先对id进行操作。先计算总行数,最后一行如果是奇数id不变,如果是偶数id减1,其余行id为奇数的让id加1,id为偶数的让id减1,最后再对id做升序操作,就可以得到结果了。
其中还有一种解法:
/* get all the even numbered rows as odd numbered rows */
SELECT s1.id - 1 as id, s1.student
FROM Seat s1
WHERE s1.id MOD 2 = 0
UNION
/* get all the odd numbered rows as even numbered rows */
SELECT s2.id + 1 as id, s2.student
FROM Seat s2
WHERE s2.id MOD 2 = 1 AND s2.id != (SELECT MAX(id) FROM Seat)
/* Just don't get the last row as we will handle it in the next UNION */
UNION
/* get the last row if odd and don't change the id value */
SELECT s3.id, s3.student
FROM Seat s3
WHERE s3.id MOD 2 = 1 AND s3.id = (SELECT MAX(id) FROM Seat)
/* Order the result by id */
ORDER BY id ASC;
思路都和第一种方法大同小异。