Question

我有使用递归和2D Ascii图像在Java中编写泛洪填充算法的任务。我写了代码并且它工作得很好但是我不确定我是否可以更简单地编写它因为我使用了太多if语句来检查一些东西，比如当前点（边缘，角落或中间）。

以下是代码：

public class AsciiShop {

public static void main(String[] args) {
    String[] img = new String[5];

    img[0] = "++++++#";
    img[1] = "+#+++##";
    img[2] = "###++#+";
    img[3] = "+#++#++";
    img[4] = "+####++";


    fill(img, 1, 2, '0');

}

public static void fill(String[] image, int x, int y, char c) {

    String[] finalImage = image;

    char startChar = finalImage[y].charAt(x);

    StringBuilder stringBuilder = new StringBuilder(finalImage[y]);
    stringBuilder.setCharAt(x, c);

    finalImage[y] = stringBuilder.toString();

    if(y>0&&x>0&&y<(finalImage.length-1)&&x<(finalImage[y].length()-1)) {
        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);
    } 
    else if(y==0&&x==0) {

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);

    }

    else if (y==0&&x==(finalImage[y].length()-1)) {
        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);
    }

    else if (y==finalImage.length&&x==(finalImage[y].length()-1)) {

        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

    }

    else if (y==finalImage.length&&x==0) {

        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);

    }

    else if (y==0&&x>0&&x<(finalImage[y].length()-1)){

        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);
    }

    else if (y==(finalImage.length-1)&&x>0&&x<(finalImage[y].length()-1)){
        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);
    }

    else if (x==0&&y>0&&y<(finalImage.length-1)) {


        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

        //Rechter Nachbar
        if(finalImage[y].charAt(x+1) == startChar)
            fill(finalImage, x+1, y, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);
    }
    else if (x==(finalImage[y].length()-1)&&y>0&&y<(finalImage.length-1)) {

        //Linker Nachbar
        if(finalImage[y].charAt(x-1) == startChar)
            fill(finalImage, x-1, y, c);

        //Nachbar oben
        if(finalImage[y-1].charAt(x) == startChar)
            fill(finalImage, x, y-1, c);

        //Nachbar unter
        if(finalImage[y+1].charAt(x) == startChar)
        fill(finalImage, x, y+1, c);
    }

    for (int i=0; i<finalImage.length; i++) {
        System.out.println(finalImage[i]);
    }

    System.out.println();

}
}

Answer 1

为什么不结合边界检查＆amp; “边界/已填充”检查并将它们移动到自己的方法中？

public class ImageFiller {
    protected String[] finalImage;    // init these in constructor
    protected int sizeX;
    protected int sizeY;
    protected char startChar;
    protected char fillChar;


    public void fill (int x, int y) {
        // ...
        visitNeighbor( x-1, y);
        visitNeighbor( x+1, y);
        visitNeighbor( x, y-1);
        visitNeighbor( x, y+1);
        // ...
    }

    protected void visitNeighbor (int x, int y) {
        if (x < 0 || x >= sizeX)
            return;
        if (y < 0 || y >= sizeY)
            return;
        if (finalImage[y].charAt(x) != startChar) {
            // Boundary or Already Filled.
            return;
        }
        // Recursive Fill;   
        //  -- or could use a queue, to avoid excessively deep recursion.
        fill( x, y);
    }
}

Answer 2

你可以循环遍历每个方向，然后将你的边界检查放在循环中。为了遍历方向，常见的技术是保持大小相等的dx和dy数组，表示每个方向的x和y的变化。对不起，我对C感觉更舒服，但语言的语法相似，所以希望它还可以。

char img[5][7], finalimg[5][7];
int h = 5, w = 7;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int numdir = 4;

void startfill(int y, int x, char c) { // Copy img into finalimg then fill
    for (int i = 0; i < h; i++) {
        for (int j = 0; j < w; j++) {
            finalimg[i][j] = img[i][j];
        }
    }
    fill(y, x, c);
}

void fill(int y, int x, char c) {
    char startchar = finalimg[y][x];
    finalimg[y][x] = c;
    for (int i = 0; i < numdir; i++) {
        int newy = y + dy[i];
        int newx = x + dx[i];
        if (newy >= 0 && newy < h && newx >= 0 && newx < w && finalimg[newy][newx] == startchar) {
            fill(newy, newx, c);
        }
    }
}

Java中的递归Floodfill

2 个答案: