Android的 - 从位置A（​​指数）检查是否现在的位置是B（指数）是对角线它在GridView布局不管接近 [英] Android - Check from position A (index) if postion B (index) is diagonal to it in a gridview layout regardless of proximity

问题描述

如果我有A和B位置的位置或行/列，检查是否B是对角线阿？

  1 2 3
 4 5 6
 7 8 9
 

我如何检查是否例如，如果5角至7？

另外，如果我检查，看看是否是4角3，它不应该得到真正的回报也？

解答从下面集成在我的处境。

我就可以从另一个函数调用

 ！isDiagonal（goodguyposition，positionadd1）公共静态布尔isDiagonal（INT A，INT B）
    {
        // X =列数
        // Y =行数
        // S =指数开始（1）
        // A =的指数
        // B =的索引        INT X = 11;
        // INT Y = 11;
        INT S = 0;
        INT AX =（S  -  A）％X，AY =（S  -  A）/ X，BX =（S  -  B）％X，用=（S  -  B）/ X;        如果（（斧== BX  -  1 ||斧== BX + 1）及及（AY ==由 -  1 || AY ==由+ 1））
        {
            返回true;
        }
        其他
        {
            返回false;
        }    }
 

解决方案

响应清洗可读性：

 公共静态布尔isAdjacentDiagonal（INT X，int类型，int类型的，INT B）{
    // X =列数
    // S =指数开始
    // A =的指数
    // B =的索引
    INT AX =（A  -  S）为％x，AY =（A  -  S）/ X，BX =（B  -  S）为％x，通过=（B  -  S）/ X;
    回报（BX ==斧 -  1 || BX == AX + 1）及和放大器; （由== AY  -  1 ||由== AY + 1）;
}
 

使用Math.abs且仅当你的索引从0开始的：

 公共静态布尔isAdjacentDiagonal（INT X，int类型的，整数B）{
    INT AX =一％×，AY = A / X，BX = B％X，由= B / X;
    返回Math.abs（斧 -  BX）== 1和;＆放大器; Math.abs（AY  - 通过）== 1;
}公共静态布尔isOneOfDiagonals（INT X，int类型的，整数B）{
    INT AX =一％×，AY = A / X，BX = B％X，由= B / X;
    返回= B和;！＆安培; Math.abs（AX  -  BX）== Math.abs（AY  - 通过）;
}
 

If I have a position or row/column for both A and B positions, check to see if B is diagonal to A?

 1 2 3
 4 5 6
 7 8 9

How do I check if for example if 5 is diagonal to 7?

Also if I check to see if 4 is diagonal 3, it shouldn't get a true return also?

Answer from below integrated in my situation.

My call on it from another function

!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;
        }

    }

解决方案

Response cleaned for readability:

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);
}

With Math.abs and only if your index starts at 0:

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);
}

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