Question

我想采用下面的矩阵并将字符TOAD重新排列为ADOT并将下面的相应列重新排列到上面移动的字符的位置，例如A移动到col 0所以现在VGD应该都在col 0等

TOAD是一个独立的阵列！我正在使用该关键字按字母顺序对martrix进行排序。

    T O A D 
    V V V X 
    D V G G 
    D F D V

我已经走到了这一步，基本上我想知道的是，如果A移动到开始我如何让下面的整个列随之移动？

public class CipherKey {

Scanner write = new Scanner (System.in);
Matrix polyCipher;
private final char[] PHRASE_KEY = { 'J', 'A', 'V', 'A'};
String[][] matrixStore = new String[PHRASE_KEY.length][PHRASE_KEY.length];
String outmrk = "|";
String inmrk = ",";
CipherKey(Matrix m) {
    polyCipher = m;
}


public void assembleMatrix()  { 

    String s = matrixFormatter(polyCipher.getTranslation());
    matrixStore = buildMatrix(s,outmrk,inmrk);
    System.out.print(printMatrix(matrixStore));
    System.out.print(sortMatrix(matrixStore));


}


public String printMatrix(String s [][]){
    char key[] = polyCipher.getKeyword().toCharArray();
    String keyOut = "";
    for(int i=0;i<key.length;i++){
        keyOut += key[i] + " ";
    }
    keyOut += "\n";
    for (int i = 0; i < s.length; i++) {
   //     keyOut += PHRASE_KEY[i] + " ";
       for (int j = 0; j < s[i].length; j++) {
            keyOut += s[i][j] + " ";
        }
        keyOut += "\n";
    }
    return keyOut.toUpperCase();
}

public static String [][] buildMatrix (String translation, String outermarker, String innermarker) {
    // outerdelim may be a group of characters
    String [] sOuter = translation.split("[" + outermarker + "]"); 
    int size = sOuter.length;
    // one dimension of the array has to be known on declaration:
    String [][] result = new String [size][size]; 
    int count = 0;
    for (String line : sOuter)
    {
        result [count] = line.split (innermarker);
        ++count;
    }
    return result;
}

以下是我遇到问题的部分：

public String sortMatrix(String a [][]){

    System.out.println("\nAlphabetically rearranged: ");
    char[] key = polyCipher.getKeyword().toCharArray();
    char[] alphaKey = polyCipher.getKeyword().toCharArray();
    String alphaOut = "";
    for(int i=0;i<alphaKey.length;i++){
        alphaOut += alphaKey[i] + " ";
    }

        char[][] x = new char[4][36];
        x[0] = new char[]{key[0], key[1], key[2], key[3]};
        x[1] = new char[]{'V', 'V', 'V', 'X'};
        x[2] = new char[]{'D', 'V', 'G', 'G'};
        x[3] = new char[]{'D', 'F', 'D', 'V'};

        String [] strArray = new String[x[0].length];

        for (int loop1 = 0; loop1 < x.length; loop1++){
            for (int loop2 = 0; loop2 < x[0].length; loop2++){
                if(strArray[loop2] == null){
                    strArray[loop2] = "";
                }
                strArray[loop2] += x[loop1][loop2];
            }
        }

        Arrays.sort(strArray);

        alphaOut += "\n";
        for (int i = 0; i < strArray.length; i++) {
       //     keyOut += PHRASE_KEY[i] + " ";
           for (int j = 0; j < s[i].length; j++) {
               alphaOut += s[i][j] + " ";
            }
           alphaOut += "\n";
        }
        alphaOut.toUpperCase();


}


public String matrixFormatter(String x){

     String resultstr = "";
      int i = 0;
      while(i < x.length()) {
        // If end of string: only add character.
        if (i == x.length() - 1) {
          resultstr += x.substring(i, i + 1);
        } else {
          if ( ((i + 1) % 4) == 0) {
            resultstr += x.substring(i, i + 1)  + "|";
          } else {
            resultstr += x.substring(i, i + 1)  + ",";
          }
        }
        i++;
      }
      return resultstr;

}

}

Answer 1

为此，您必须跟踪移动并相应地移动其他元素。但是看看你的程序，我可以看到你依靠java API进行排序。

如果你想继续这样做，我可以给你一个适合你的小技巧。

第1步：定义数组：

    char[][] x = new char[4][4];
    x[0] = new char[]{'T', 'O', 'A', 'D'};
    x[1] = new char[]{'V', 'V', 'V', 'X'};
    x[2] = new char[]{'D', 'V', 'G', 'G'};
    x[3] = new char[]{'D', 'F', 'D', 'V'};

步骤2：将其转换为垂直连接字符的单维字符串数组。

    String [] strArray = new String[x[0].length];

    for (int loop1 = 0; loop1 < x.length; loop1++){
        for (int loop2 = 0; loop2 < x[0].length; loop2++){
            if(strArray[loop2] == null){
                strArray[loop2] = "";
            }
            strArray[loop2] += x[loop1][loop2];
        }
    }

第3步：对字符串数组进行排序

    Arrays.sort(strArray);

第4步：将其转换回char数组

    for (int loop1 = 0; loop1 < strArray.length; loop1++){
        for (int loop2 = 0; loop2 < strArray[loop1].length(); loop2++){
            x[loop2][loop1] = strArray[loop1].charAt(loop2);
        }
    }

这应该可以解决问题。

修改假设您必须根据char [] alpha的位置切换String a [] []的位置，您必须将其转换为char [] []数组，就像我在下面所做的那样：

char[] alpha = {'T', 'O', 'A', 'D'}; String a [][] = {{"V", "V", "V", "X"},{"D", "V", "G", "G"},{"D", "F", "D", "V"}}; char[][] x = new char[a.length+1][]; x[0] = alpha; for(int loop1 = 0; loop1 < a.length; loop1++){ x[loop1+1] = new char[a[loop1].length]; for (int loop2 = 0; loop2 < a[loop1].length; loop2++){ x[loop1+1][loop2] = a[loop1][loop2].charAt(0); } }

这会将您的两个数组转换为一个二维数组，您可以在其上运行其余步骤：

char[][] x = new char[4][4]; x[0] = new char[]{'T', 'O', 'A', 'D'}; x[1] = new char[]{'V', 'V', 'V', 'X'}; x[2] = new char[]{'D', 'V', 'G', 'G'}; x[3] = new char[]{'D', 'F', 'D', 'V'};

Answer 2

由于矩阵是数组（列）的数组（行），因此最好首先转置数据。然后可以有效地完成切换行（较早的列）：

void switchRows(String[][] a, int i, int j) {
    String[] temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}

转置意味着a [i] [j]应该被填充为旧a [j] [i]。

也许您可以考虑使用char代替String。

Answer 3

转置数据，使上面显示的行成为列，反之亦然。由于二维数组是一个数组数组，因此您希望每个子数组包含一列中的所有字符。然后你可以编写一个函数，用每个子数组的第一个索引的内容对子数组进行排序。

    char[][] unsorted = new char[4][4];
    unsorted[0] = new char[]{'T', 'V', 'D', 'D'};
    unsorted[1] = new char[]{'O', 'V', 'V', 'F'};
    unsorted[2] = new char[]{'A', 'V', 'G', 'D'};
    unsorted[3] = new char[]{'D', 'X', 'G', 'V'};

    // Now sort unsorted[0] through unsorted[3] by the entries at unsorted[i][0]

您可以编写一个简单的Comparator类，并使用public static <T> void sort(T[] a, Comparator<? super T> c)中的java.util.Arrays方法。或者写一个简单的插入排序。

有没有办法在2d数组矩阵中移动列？

3 个答案: