螺旋顺序的二维数组 [英] 2d Array in Spiral Order

问题描述

我正在尝试按螺旋顺序填充数组.到目前为止，我可以按螺旋顺序打印数组，但是有没有办法修改数组，以便我可以按螺旋顺序填充它，然后只打印数组?我希望它像倒计时一样按降序排列.请帮忙！

I'm trying to fill an array in spiral order. So far, I can print the array in spiral order, but is there a way to modify the array so that i can fill it in spiral order and then just print the array? I'd like it to go in decreasing order like a countdown. Please help!

public class Spiral {
  public static void main(int m, int n) { 

    // create m by n array of integers 1 through m*n
    int[][] values = new int[m][n];
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            values[i][j] = 1 + (m*n)*i + j;

    // spiral
    for (int i = (m*n)-1, j = 0; i > 0; i--, j++) {
          for (int k = j; k < i; k++) System.out.println(values[j][k]);
          for (int k = j; k < i; k++) System.out.println(values[k][i]);
          for (int k = i; k > j; k--) System.out.println(values[i][k]);
          for (int k = i; k > j; k--) System.out.println(values[k][j]);
    }
  }
}

推荐答案

下面的函数是一个大小为 N × N 的方阵，其中包含从 1 到 N * N 的整数，按螺旋顺序从左上角开始顺时针方向方向.

Below function is a square matrix with a size N × N containing integers from 1 to N * N in a spiral order, starting from top-left and in clockwise direction.

int[][] spiralNumbers(int n) {
int[][] matrix = new int[n][n];
for (int step = 0, a = 0, size; step < n/2; step++) {
    size = (n - step * 2 - 1);
    for (int i = 0, chunk, chunkIndex, chunkOffset; i < 4 * size; i++) {
        chunk = i / size;
        chunkIndex = i % size;
        chunkOffset = n - step - 1;
        switch (chunk) {
            case 0:
                matrix[step][chunkIndex + step] = a+1;
                break;
            case 1:
                matrix[chunkIndex + step][chunkOffset] = a+1;
                break;
            case 2:
                matrix[chunkOffset][chunkOffset - chunkIndex] = a+1;
                break;
            case 3:
                matrix[chunkOffset - chunkIndex][step] = a+1;
                break;
            default:
                throw new IndexOutOfBoundsException();
        }
        a++;
    }
    if (n % 2 == 1) {
        matrix[n/2][n/2] = n * n;
    }
}
return matrix; 
}

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