java - 2D 配列 Java の比較

Question

指定された2番目の配列が最初の配列のどこに配置されるかを調べようとしていますが、文字列に変換してそれらがどこにあるかを調べることができないようです:

public static void main(String[] args) {
    char i [][] = new char[][]{
            {'d', 's', 'l', 'e', 'i', 'g', 'h', 'e', 'i', 'j', 'a', 's', 'l', 'd', 'k', 'j'},
            {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'W', 'w', 'Z', 'Z', 'Z', 'W', '1', 'l', 'k'},
            {'h', 'i', 'j', 'k', 'l', 'm', 'n', 'Z', 'A', 'a', 'Z', 'a', 'Z', '2', 'i', 'n'},
            {'o', 'p', 'q', 'r', 's', 't', 'u', 'Z', 'Z', 'L', 'l', 'Z', 'Z', '3', 'i', 'v'},
            {'v', 'w', 'x', 'y', 'z', '1', '2', 'Z', 'd', 'Z', 'D', 'd', 'Z', '4', 'q', 'i'},
            {'3', '4', '5', '6', '7', '8', '9', 'o', 'Z', 'Z', 'o', 'O', 'Z', '5', 'b', 'v'},
            {'k', 'e', '8', '7', '8', '4', 'j', 'f', 'l', 'k', 'a', 'v', '8', '8', 'h', 'j'}
    };

    char w [][] = new char[][]{
            {'W', 'w', '.', '.', '.', 'W', '1'},
            {'.', 'A', 'a', '.', 'a', '.', '2'},
            {'.', '.', 'L', 'l', '.', '.', '3'},
            {'.', 'd', '.', 'D', 'd', '.', '4'},
            {'o', '.', '.', 'o', 'O', '.', '5'}
    };

    find(i, w);
}

    public static int[] find(char [][] image, char [][] waldo) {
    for (int i = 0; i < waldo.length; i++) {
        char[] largerCharArray= large[i];
        String largerString = new String(largerCharArray);

        //used for debug purposes
        char[] array = largerCharArray;

        char [] smallCharArray = small[i];
        String smallString = new String(smallCharArray);

        char[] array1 = smallCharArray;

        //beginning comparison
        if (largerString.indexOf(smallString) >= 0) {

           int temp = largerString.indexOf(smallString);
        }

        //debug purposes
        System.out.println(largerString.indexOf(smallString));
        System.out.println(Arrays.toString(array));
        System.out.println(Arrays.toString(array1));
    }
    //for debug purposes
    return null;
}

Answer 1

完成したばかりのこのコードを試してみてください。バグがあるかもしれませんが、改善を試みます。文字列は使用しません。

private static int[] find(char[][] image, char[][] waldo) {
    int row = -1;
    int column = -1;
    char first_waldo = waldo[0][0]; // first char of waldo
    boolean continue1 = true; // this is used to avoid looping when the indexes tried are not a possible answer
    int size = waldo.length * waldo[0].length; // number of elements of waldo
    int cont = 0;

    for (int i = 0; i < image.length; i++) {
        for (int j = 0; j < image[i].length; j++) { // Looping over chars in image array
            char current = image[i][j]; // Current char of image array
            if (current == first_waldo) { // If true, this indexes (i, j) are a possible answer
                row = i; // Current image row
                column = j; // Current image column
                cont = 0; // Count of how many chars are matched
                for (int k = row; k < row + waldo.length; k++) {
                    for (int l = column; l < column + waldo[0].length; l++) { // Looping over
                                                       // the possible matching array in image
//                          System.out.println("Comparing " + image[k][l] + " with " + waldo[k - row][l - column]);
//                          System.out.println(k + " " + l);
                        if (waldo[k - row][l - column] == '.') { // If its a point, count as matching characters
                            cont++;
                            continue; 
                        }
                        if (image[k][l] != waldo[k - row][l - column]) { // If chars are different, it's not the answer
                            row = -1;
                            column = -1;
                            continue1 = false; // So it doesn't continue looping
                            break;
                        } else {
                            cont++; // If they are equals, count as matching characters
                        }
                    }
                    if (continue1 == false) { // To avoid overlooping when it's not the answer
                        continue1 = true; // Reset value
                        break;
                    }
                }
                if (cont == size) { // Is the number of matched charactes equal to the size of waldo?
                    int[] res = {row, column};
                    return res; // Return indexes
                }

            } else {
                row = -1;
                column = -1;
            }

        }
    }
    int[] res = {-1, -1};
    //      System.out.println("\n" + row +" , " +  column);
    return res; // Not answer

}

Answer 2

私が理解しているように、タスクは一致する部分行列を見つけることです。ドットをワイルドカードに置き換えると、意味があります。したがって、小さい方の行列は、1 から数えて (行 = 2、列 = 8) の位置にあります。

サブマトリックスを検索するコードは次のとおりです。正規表現を使用して、行を文字列として一致させました。

public static int[] find(char [][] large, char [][] small) {
    outer: for (int i = 0; i < large.length - small.length; i++) {

        int column = -1;

        for (int j = 0; j < small.length ; j++) {
            String largerString = new String(large[i + j]);

            String smallString = new String(small[j]);

            Matcher matcher = Pattern.compile(smallString).matcher(largerString);

            if(!matcher.find()){
                continue outer;
            }

            int thisColumn = matcher.start();

            if(column != -1 && column != thisColumn){
                continue outer;
            }

            column = thisColumn;
        }

        System.out.printf("found it: %d, %d!", i, column);
    }

    //for debug purposes
    return null;
}