README.md

comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0626.Exchange%20Seats/README.md	数据库

626. 换座位

English Version

题目描述

表: Seat

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| student     | varchar |
+-------------+---------+
id 是该表的主键（唯一值）列。
该表的每一行都表示学生的姓名和 ID。
id 是一个连续的增量。

 

编写解决方案来交换每两个连续的学生的座位号。如果学生的数量是奇数，则最后一个学生的id不交换。

按 id 升序 返回结果表。

查询结果格式如下所示。

 

示例 1:

输入: 
Seat 表:
+----+---------+
| id | student |
+----+---------+
| 1  | Abbot   |
| 2  | Doris   |
| 3  | Emerson |
| 4  | Green   |
| 5  | Jeames  |
+----+---------+
输出: 
+----+---------+
| id | student |
+----+---------+
| 1  | Doris   |
| 2  | Abbot   |
| 3  | Green   |
| 4  | Emerson |
| 5  | Jeames  |
+----+---------+
解释:
请注意，如果学生人数为奇数，则不需要更换最后一名学生的座位。

解法

方法一

MySQL

# Write your MySQL query statement below
SELECT s1.id, COALESCE(s2.student, s1.student) AS student
FROM
    Seat AS s1
    LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
ORDER BY 1;

方法二

MySQL

# Write your MySQL query statement below
SELECT
    id + (
        CASE
            WHEN id % 2 = 1
            AND id != (SELECT MAX(id) FROM Seat) THEN 1
            WHEN id % 2 = 0 THEN -1
            ELSE 0
        END
    ) AS id,
    student
FROM Seat
ORDER BY 1;

方法三

MySQL

# Write your MySQL query statement below
SELECT
    RANK() OVER (ORDER BY (id - 1) ^ 1) AS id,
    student
FROM Seat;

方法四

MySQL

# Write your MySQL query statement below
SELECT
    CASE
        WHEN id & 1 = 0 THEN id - 1
        WHEN ROW_NUMBER() OVER (ORDER BY id) != COUNT(id) OVER () THEN id + 1
        ELSE id
    END AS id,
    student
FROM Seat
ORDER BY 1;