代码如下：

Question

我想从以下矩阵中打印对角线

XVSOKG
WTPLHD
UQMIEB
RNJFCA

从右下角开始，因此输出应为A，B，C，D等。我写下了元素和跟随这样的索引 A [3,5]，B [2,5]，C [3,4]，D [1,5]，E [2,4]，F [3,3]等，但我找不到任何从这个模式。我怎么想这样做？

Answer 1

事实上，模式非常简单，问题是您需要了解边界。我做了一个能够完成你所寻找的功能（如果我没有误解，你发布的矩阵你想要打印字母表）。不幸的是，我没有在C＃中做任何设置，所以我用Java做了，希望你能简化它并轻松移植它。

我测试了不同大小的矩阵，但是......你知道。如果可以，请进行测试。

使用以下矩阵：

String[][] matrix = new String[][]{
            {"X","V","S","O","K","G"},
            {"W","T","P","L","H","D"},
            {"U","Q","M","I","E","B"},
            {"R","N","J","F","C","A"}
        };

结果是：

3 - 5 = A
2 - 5 = B
3 - 4 = C
1 - 5 = D
2 - 4 = E
3 - 3 = F
0 - 5 = G
1 - 4 = H
2 - 3 = I
3 - 2 = J
0 - 4 = K
1 - 3 = L
2 - 2 = M
3 - 1 = N
0 - 3 = O
1 - 2 = P
2 - 1 = Q
3 - 0 = R
0 - 2 = S
1 - 1 = T
2 - 0 = U
0 - 1 = V
1 - 0 = W
0 - 0 = X

代码如下：

（这看起来很长，但主要是评论）

/**
 * Prints the diagonals of a matrix from bottom right to top left.
 * 
 * @param matrix the matrix to print
 */
public static void printDiagonals(String[][] matrix){
    // extract the width and the height of the matrix
    int width = matrix[0].length;
    int height = matrix.length;

    /**
     * these are the pointers that go through the matrix, we inicialize it
     * in the bottom right corner 
     */
    int xIndex = width - 1;
    int yIndex = height - 1;
    /**
     * these are auxiliar pointers to let me trace in what diagonal is the
     * next, they are optional but help making the code more clear; I 
     * initialize it with the position above the bottom right corner
     */
    int startingX = xIndex;
    int startingY = yIndex - 1;

    //while we have not reach the top left corner
    while (xIndex >= 0 && yIndex >= 0){
        // print the current index and the value, here change it with watever you need
        System.out.println(yIndex + " - " + xIndex + " = " + matrix[yIndex][xIndex]);
        /**
         * as we are moving diagonally to the bottom left we achieve this by
         * decreasing the x index and increasing the Y index
         */
        xIndex--;
        yIndex++;
        //if we hit one of the borders
        if (yIndex == height || xIndex < 0){
            //move the indices to the next diagonal
            yIndex = startingY;
            xIndex = startingX;
            //then we adjust the next diagonal
            if (startingY > 0){
                startingY--;
            } else {
                startingX--;
            }
        }
    }
}

对角打印数组

1 个答案:

代码如下：