螺旋顺序的2d数组

Question

我正在尝试以螺旋顺序填充数组。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顺序递减。请帮忙！

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]);
    }
  }
}

Answer 1

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);

}

Answer 2

下面的函数是一个大小为N×N的方矩阵，其中包含从1到N * N的螺旋形整数，从左上角开始沿顺时针方向旋转。

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; 
}

Answer 3

如果您已经找到了执行读取的代码（用于打印），那么您肯定可以使用相同的逻辑来修改它以进行写入吗？

如果您希望矩阵中的每个单元格都包含其“序列号”，向后计数，这样的事情应该有效，假设您的访问逻辑是正确的：

for (int i = (m*n)-1, j = 0, index = m * n; i > 0; i--, j++) {
      for (int k = j; k < i; k++) values[j][j] = index--;
      for (int k = j; k < i; k++) values[k][i] = index--;
      for (int k = i; k > j; k--) values[i][k] = index--;
      for (int k = i; k > j; k--) values[k][j] = index--;
}

Answer 4

效率不高，但应该有效：g是数组。我也使用异常来控制逻辑。

public static void spiralFill()
{
    x = (g.length-1)/2;
    y = (g[0].length-1)/2;

    try
    {
        while(true)
        {
             east();
             south();
             step++;
             west();
             north();
             step++;        
        }
    }
    catch(ArrayIndexOutOfBoundsException e)
    {

    }
}

public static void east()
{
for(int i = 0; i < step; i++)
{
        g[x][y] = count;
        count++;
    x++;
    }
}
public static void south()
{
    for(int i = 0; i < step; i++)
    {
        g[x][y] = count;
        count++;
        y--;
    }
}   
public static void west()
{
    for(int i = 0; i < step; i++)
    {
         g[x][y] = count;
         count++;
         x--;
    }
}   
public static void north()
{
   for(int i = 0; i < step; i++)
   {
       g[x][y] = count;
       count++;
       y++;
   }
}   

Answer 5

以下是一些变体：

public class SpiralMatrix {

    private static int[][] createSpiralMatrix(int size) {
        int[][] matrix = new int[size][size];

        int row = 0, col = -1;
        int value = 1;

        boolean horizontal = true;
        boolean increasing = true;
        boolean finish = false;

        while(!finish) {

            finish = true;
            if (horizontal && increasing) {
                while(tryAndSet(matrix, row, col + 1, value)) {
                    finish = false;
                    col++;
                    value++;
                }
            } else if (horizontal && !increasing) {
                while(tryAndSet(matrix, row, col - 1, value)) {
                    finish = false;
                    col--;
                    value++;
                }
            } else if (!horizontal && increasing) {
                while(tryAndSet(matrix, row + 1, col, value)) {
                    finish = false;
                    row++;
                    value++;
                }
            } else {
                while(tryAndSet(matrix, row - 1, col, value)) {
                    finish = false;
                    row--;
                    value++;
                }
            }

            if (!horizontal) {
                increasing = !increasing;
            }
            horizontal = !horizontal;
        }

        return matrix;
    }


    private static boolean tryAndSet(int[][] matrix, int row, int col, int value) {
        if (row < 0 || col < 0 || row >= matrix.length || col >= matrix[row].length || matrix[row][col] != 0) {
            return false;
        }
        matrix[row][col] = value;
        return true;
    }

    private static void printMatrix(int[][] matrix) {
        for (int i = 0; i < matrix.length; i++) {
            for(int j = 0; j < matrix[i].length; j++) {
                System.out.print("\t" + matrix[i][j]);
            }
            System.out.println();
        }
    }


    public static void main(String[] args) {
        try {
            int[][] spiralMatrix = createSpiralMatrix(40);
            printMatrix(spiralMatrix);
        } catch (Throwable th) {
            th.printStackTrace();
        }
    }

Answer 6

public ArrayList<Integer> spiralOrder(final List<ArrayList<Integer>> a) {
     ArrayList<Integer> result = new ArrayList<Integer>();
      int m = a.get(0).size();
      int n = a.size();
      if(m>1 && n>1){
         int loopCounter = (n > m)  ? m*2 : n *2 -1 ;
         int opr=1;
         int i=0,j=0;
         int opA=m,opB=n,opC=0,opD=1;
         for(int k=0;k < loopCounter ;k++){
              if(opr == 1){
                 int counter =0;
                 while(counter < opA){
                    System.out.print(a.get(i).get(j)+ ";");
                    result.add(a.get(i).get(j));
                    counter++;

                    if(j != opA-1){
                        j++;
                    }
                    else{
                        break;
                    }
                 }
                 opr =2;
                 continue;
             }
             if(opr == 2){
                 i++;
                 int counter =1;
                 while(counter < opB){
                     System.out.print(a.get(i).get(j)+ ";");
                    result.add(a.get(i).get(j));
                    counter++;

                   if( i != opB-1){
                     i++;
                   }
                   else{
                       break;
                   }
                 }
                 opr =3;
                 continue;
             }
             if(opr == 3){
                 j--;
                 int counter =j;
                 while(counter >= opC){
                     System.out.print(a.get(i).get(j)+ ";");
                    result.add(a.get(i).get(j));
                    counter --;
                     if(j != opC){
                     j--;
                    }
                    else{
                        break;
                    }
                  }
                 opr =4;
                 continue;
             }
             if(opr == 4){
                 i--;
                 int counter = i;
                 while(counter >= opD){
                     System.out.print(a.get(i).get(j)+ ";");
                    result.add(a.get(i).get(j));
                    counter --;
                    if(i != opD){
                     i--;
                    }else{
                        break;
                    }
                 }
                 opr =1;
                 j++;
                 opA = opA -1;
                 opB = opB -1;
                 opC= opC +1;
                 opD = opD+1;
                 continue;
             }
         }

     }

         else if(n ==1){
              for(int k=0;k < a.get(0).size();k++){
                  result.add(a.get(0).get(k));
              }
         }

       else if(m==1 && n==1){
             result.add(a.get(0).get(0));
         }

     // Populate result;
     return result;

}