我试图建立一个跟踪方法，来跟踪图像的轮廓的坐标在2dArray的形式

Question

对于一所学校PROJEKT我试图建立一个跟踪方法。它应该追查基于来自边缘检测方法的阵列形状的轮廓的坐标。我确信轮廓只有一个像素厚。因为它是一个PROJEKT我不能使用任何花哨的招数从libary。

到目前为止，我使用的是嵌套的for循环找到形状的开始。然后，我使用这样的形状：

p9  p2  p3
p8  p1  p4
p7  p6  p5

P1是我在数组中当前的位置，然后我在寻找P2-P9下一个位置。

到目前为止的代码如下所示：

private static String traceCoordinates(int o[][])
{

    String useless = null; 
    //int[][] n = new int[m[0].length][m.length];
    //int[][] o = new int[n.length][n[0].length];

   // n = transpose(m);
    //o = thinning(n);

    int hits = 0;
    System.out.print("        ");
    for (int a = 0; a < 1; a++)
    {
        for (int b = 0; b < o[0].length; b++)
        {
            String c = String.format("%3s", b);
            System.out.print(c + " ");
        }System.out.println("");
    }

    for (int y = 0; y < o.length; y++)
    {
        //String e = String.format("%3s", y).replace(" ", "0");
        //System.out.print("row" + e + " ");
        for (int x = 0 ; x < o[0].length ; x++)
        {
            if (o[y][x] == 0)
            {
                int xEnd = x;
                int yEnd = y;

                int a = y + 1;
                int b = y - 1;
                int c = x + 1;
                int d = x - 1;

                /*
                Her findes værdien af de 8 nabopixler på baggrund af deres indbyrdes index.
                */
                int p2 = o[b][x];
                int p3 = o[b][c];
                int p4 = o[y][c];
                int p5 = o[a][c];
                int p6 = o[a][x];
                int p7 = o[a][d];
                int p8 = o[y][d];
                int p9 = o[b][d];

                do
                {

                    System.out.println(y + ";" + x + " " + "begin");
                    if (p2 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        y--;
                        System.out.print("Next point: ");
                        System.out.println("2; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p3 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                       o[y][x] = 66;
                        y--;
                        x++;
                        System.out.print("Next point: ");
                        System.out.println("3; " + pntC(x, y) + " ");;
                        hits++;
                    }
                    else if (p4 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        x++;
                        System.out.print("Next point: ");
                        System.out.println("4; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p5 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        x++;
                        y++;
                        System.out.print("Next point: ");
                        System.out.println("5; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p6 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        y++;
                        System.out.print("Next point: ");
                        System.out.println("6; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p7 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        x--;
                        y++;
                        System.out.print("Next point: ");
                        System.out.println("7; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p8 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        x--;
                        System.out.print("Next point: ");
                        System.out.println("8; " + pntC(x, y) + " ");
                        hits++;
                    }
                    else if (p9 == 0 && hits < 1)
                    {
                        System.out.print("This point: ");
                        System.out.print(pntC(x, y) + " ");
                        o[y][x] = 66;
                        x--;
                        y--;
                        System.out.print("Next point: ");
                        System.out.println("9; " + pntC(x, y) + " ");
                        hits++;
                    }
                    System.out.println(y + ";" + x + " " + "end");
                    hits = 0;

                }while(o[y][x] == 0);   
                System.out.println("loop\n");



            }

        }
        System.out.println("");
    }System.out.println(hits);
    for( int row = 0 ; row < o.length ; row++)
    {//start row for loop
        String e = String.format("%3s", row);
        System.out.print("row" + e + " ");
        for (int column = 0 ; column < o[0].length ; column++)
        {// start column for loop
                           if (o[row][column] == 255)
                            {
                                       System.out.print("___|");
                            }
                           else
                           {
                               System.out.print("   " + o[row][column]);
                           }
        }// end colum for loop

        System.out.println(" end row");
    }// end row for loop
return useless;    
}

当我做，跟踪是通过一个StringBuilder作为一个字符串，因此没用变量返回。

我的问题是，我的方法只能在一个方向上检测线。当有在那个方向没有更多的像素，它跳出循环。

我练以下阵列：

int[][] A = new int[][]
    {
        {1,  2,   3,   4,   5,   6,   7,   8,   9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21, 22},
        {2,  0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 2},
        {3,  255, 0,   0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 2},
        {4,  255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 2},
        {5,  255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 2},
        {6,  255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 2},
        {7,  255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 2},
        {8,  255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 2},
        {9,  255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {10, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {11, 255, 255, 255, 255, 255, 255, 255, 255, 255, 0, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {12, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {13, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {14, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {15, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {16, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {17, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {18, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {19, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {20, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {21, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 2},
        {1,  2,   3,   4,   5,   6,   7,   8,   9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21, 22}
    };

我想看到返回的是这样的：

1;1 2;2 2;3 3;4 4;4 5;5 6;6 7;7 8;8 9;9 10;10 9;11 8;12 7;13 6;14 5;15 4;16 3;17 2;18 1;19

以该顺序。

该方法的实际输出现在的问题是：

1;1 begin

这一点：001/001下一点：5; 002/002 2; 2端2; 2开始这一点：002/002下一点：5; 003/003 3; 3端环

3; 4开始这一点：004/003下一点：6; 004/004 4; 4端部4; 4开始这一点：004/004下一点：6; 004/005 5; 4结束循环

5; 5开始这一点：005/005下一点：5; 006/006 6; 6端6; 6开始这一点：006/006下一点：5; 007/007 7; 7端7; 7开始这一点：007/007下一点：5; 008/008 8; 8端8; 8开始这一点：008/008下一点：5; 009/009 9; 9端9; 9开始这一点：009/009下一点：5; 010/010 10; 10端部10; 10开始这一点：010/010下一点：5; 011/011 11; 11结束循环

我怎样才能使这项工作？

Answer 1

对于大纲之后，知道当前像素是不够的。您还需要从它达到的方向。然后，你开始由当前像素（顺时针或逆时针）从目前的方向四处开始旋转寻找下一个像素。

因此，你的表应该有8×256项。或者，你可以尝试反过来8个邻国（实际上是6足矣）。不同空间/时间折衷是可能的。

为了开始遍历，你可以看看由一个空白后跟一个像素包围的像素的配置。停止标准是有点棘手。为了确保正确无误，当你回到初始像素，准备继续在同一个方向，你应该停下来。