Android - 从位置 A(索引)检查位置 B(索引)是否在 GridView 布局中与它成对角线，而不管是否接近

Question

如果我有一个位置或行/列同时用于 A 和 B 位置，请检查 B 是否与 A 成对角线？

 1 2 3
 4 5 6
 7 8 9

例如，我如何检查 5 是否与 7 成对角线？

此外，如果我检查 4 是否是对角线 3，它不应该也得到真正的返回吗？

下面的回答结合了我的情况。

我从另一个函数调用它

!isDiagonal(goodguyposition, positionadd1)

public static boolean isDiagonal(int a, int b) 
    {
        // x = number of columns
        // y = number of rows
        // s = index start (1)
        // a = index of a
        // b = index of b

        int x = 11;
        //int y = 11;
        int s = 0;
        int ax = (s - a) % x, ay = (s - a) / x, bx = (s - b) % x, by = (s - b) / x;

        if ((ax == bx - 1 || ax == bx + 1) && (ay == by - 1 || ay == by + 1))
        {
            return true;
        }
        else
        {
            return false;
        }

    }

Answer 1

为提高可读性而清理的响应:

public static boolean isAdjacentDiagonal(int x, int s, int a, int b) {
    // x = number of columns
    // s = index start
    // a = index of a
    // b = index of b
    int ax = (a - s) % x, ay = (a - s) / x, bx = (b - s) % x, by = (b - s) / x;
    return (bx == ax - 1 || bx == ax + 1) && (by == ay - 1 || by == ay + 1);
}

使用 Math.abs 且仅当您的索引从 0 开始时:

public static boolean isAdjacentDiagonal(int x, int a, int b) {
    int ax = a % x, ay = a / x, bx = b % x, by = b / x;
    return Math.abs(ax - bx) == 1 && Math.abs(ay - by) == 1;
}

public static boolean isOneOfDiagonals(int x, int a, int b) {
    int ax = a % x, ay = a / x, bx = b % x, by = b / x;
    return a != b && Math.abs(ax - bx) == Math.abs(ay - by);
}