Leetcode613.直线上的最近距离（简单）

问题
point 表包含了一些点的x坐标值，坐标值均为整数。

写一条查询语句计算这些点中的最短距离。

x
-1
0
2

最短距离明显是1，是-1和0之间的距离。所以输出如下：

shortest
1

注意: 每个点都是唯一的， point 表中没有重复记录。

生成数据

CREATE TABLE POINT(X INT);

INSERT INTO POINT VALUE(-1),(0),(2),(4);

SELECT * FROM POINT;

解答
想法是 对x进行排序 就可以得到类似于数轴的关系 既然要求最近距离就可以只求相邻两个值的差值 再找最小即可
但是这样取需要引入id来确认他们是否相邻
尝试一下

SELECT X ,@num:=@num+1 AS num
FROM POINT, (SELECT @num:=0) AS tmp
ORDER BY X DESC;

把相邻两者连接 num1和num要做去区分

SELECT *
FROM (SELECT X ,@num:=@num+1 AS num
FROM POINT, (SELECT @num:=0) AS tmp
ORDER BY X DESC) A1,
(SELECT X ,@num1:=@num1+1 AS num1
FROM POINT, (SELECT @num1:=0) AS tmp
ORDER BY X DESC) A2
WHERE A1.num = A2.num1-1;

选出相邻之差的最大值即可

SELECT MAX(A1.X -A2.X) AS shortest
FROM (SELECT X ,@num:=@num+1 AS num
FROM POINT, (SELECT @num:=0) AS tmp
ORDER BY X DESC) A1,
(SELECT X ,@num1:=@num1+1 AS num1
FROM POINT, (SELECT @num1:=0) AS tmp
ORDER BY X DESC) A2
WHERE A1.num = A2.num1-1;

别的解答
构造偏序关系

select min(P2.x-P1.x) as `shortest`
from Point as P1 join Point as P2
    on(P1.x < P2.x)

也可以把不等的两个数值拿来做差取绝对值的最大值

select min(abs(P2.x-P1.x)) as `shortest`
from Point as P1 join Point as P2
    on(P1.x <> P2.x)