在螺旋的次序二维数组

问题描述

我试图填补螺旋的次序数组。到目前为止，我可以打印在螺旋的顺序排列，但有没有办法修改数组，这样我可以在螺旋为了填补它，然后只打印数组？我想它去按递减顺序就像一个倒计时。请帮帮忙！

 公共类螺旋{
  公共静态无效的主要（INT男，INT N）{

    //通过M * N新建米×1的整数n个阵列
    INT [] []值=新INT [M] [N];
    的for（int i = 0; I＆LT;米;我++）
        为（诠释J = 0; J＆n种; J ++）
            值[I] [J] = 1 +（M * N）* I + J;

    // 螺旋
    对（INT I =（M * N）-1，J = 0; I＆≠0; I--，J ++）{
          对于（时int k = j的; K＆LT;我; k ++）的System.out.println（价值[J] [K]）;
          对于（时int k = j的; K＆LT;我; k ++）的System.out.println（值[k]的[I]）;
          对于（INT K = 1; K＆GT; D]; K--）的System.out.println（值[I] [K]）;
          对于（INT K = 1; K＆GT; D]; K--）的System.out.println（值[k]的[J]）;
    }
  }
}
 

解决方案

  INT包括maxValue = target.length *目标[0] .length;

私人INT [] [] generateMatrix（INT [] []的目标，INT层面，诠释currentVal）{
    //总是从每层中左下角启动
    INT W =水平;
    INT H = target.length  -   -  1;

    //填写底线
    INT I = 0;
    对（我= W; I＆lt;目标[0] .length  - 水平和放大器;＆安培; currentVal＆LT; = maxValue（最大值）;我++）{
        目标[H] [I] = currentVal ++;
    }

    W =靶[0] .length  -   -  1;

    INT J = 0;
    //填写右边线
    对于（，J = H  -  1; J＆GT; =水平和放大器;＆安培; currentVal＆LT; = maxValue（最大值）; j--）{
        目标[J] [W] = currentVal ++;
    }

    H =水平;

    //填补上述行
    对（我= W  -​​  1; I＆GT; =水平和放大器;＆安培; currentVal＆LT; = maxValue（最大值）;我 - ）{
        目标[H] [I] = currentVal ++;
    }
    W =水平;

    //填补了左线
    为（J = H + 1; J＆LT; target.length  - 级 -  1安培;＆安培; currentVal＆LT; = maxValue（最大值），J ++）{
        目标[J] [W] = currentVal ++;
    }

    如果（currentVal＆GT;包括maxValue）
        将目标;
    返回generateMatrix（目标，++级，currentVal）;

}
 

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]);
    }
  }
}

解决方案

int maxValue = target.length * target[0].length;

private int[][] generateMatrix(int[][] target, int level, int currentVal) {
    // always start from lower left corner in each layer
    int w = level;
    int h = target.length - level - 1;

    // fill the bottom line
    int i = 0;
    for (i = w; i < target[0].length - level && currentVal <= maxValue; i++) {
        target[h][i] = currentVal++;
    }

    w = target[0].length - level - 1;

    int j = 0;
    // fill the right line
    for (j = h - 1; j >= level && currentVal <= maxValue; j--) {
        target[j][w] = currentVal++;
    }

    h = level;

    // fill the above line
    for (i = w - 1; i >= level && currentVal <= maxValue; i--) {
        target[h][i] = currentVal++;
    }
    w = level;

    // fill the left line
    for (j = h + 1; j < target.length - level - 1 && currentVal <= maxValue; j++) {
        target[j][w] = currentVal++;
    }

    if (currentVal > maxValue)
        return target;
    return generateMatrix(target, ++level, currentVal);

}

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