Question

我需要用Java编写算法来验证密码字段。在所有基本控件之间（最小长度，lumhezza massima，字母数字...），我必须检查密码是否不是由键盘上的顺序字符组成的（例如 QWERTY，YTREWQ，ASDFGH 等）。密码的最小长度为8个字符。密码在4位数字后无效我无法在数据库中使用“禁止”字符串，关于现有算法的任何提示，或者是否有图书馆已经在进行这种检查，来创建“字典表”？

Answer 1

嗯...我想到一件事。

您可以尝试构建一个字母矩阵：

-----------------------
| q  w  e  r  t  y  ... 
| a  s  d  f  g  h  ... etc.
| z  x  c  v  b  n  ... 
-----------------------

给每个字母一个索引。例如，“ q”具有[1,1]，“ w”具有[2，1]，“ v”具有[4,3]，依此类推。

现在，给定一个字符串，找到每个字母之间的距离并求和。

示例：
给定字符串“ sdgh”：

s-d, distance is 1
d-g, distance is 2
g-h, distance is 1

Total distance is: 4

现在您的比率为4/4。（四个字母，总距离为四个）。
您的安全要求可以吗？回答这个问题，您就完成了。

Answer 2

好的，这是我的解决方案，对于水平检查似乎效果很好。如果距离> 4，则表示密码无效。有任何改进的提示吗？我知道它有点硬编码，但目前我可以通过它。

    @Test
public void runCheckDigit() {

    checkdigit("cnqwerty13");

}

public void checkdigit(String password) {

    int x = 4;
    int y = 10;

    String[][] keyboard = new String[x][y];


    // first row
    keyboard[0][0] = "1";
    keyboard[0][1] = "2";
    keyboard[0][2] = "3";
    keyboard[0][3] = "4";
    keyboard[0][4] = "5";
    keyboard[0][5] = "6";
    keyboard[0][6] = "7";
    keyboard[0][7] = "8";
    keyboard[0][8] = "9";
    keyboard[0][9] = "0";

    // second row
    keyboard[1][0] = "q";
    keyboard[1][1] = "w";
    keyboard[1][2] = "e";
    keyboard[1][3] = "r";
    keyboard[1][4] = "t";
    keyboard[1][5] = "y";
    keyboard[1][6] = "u";
    keyboard[1][7] = "i";
    keyboard[1][8] = "o";
    keyboard[1][9] = "p";
    // TERZA RIGA
    keyboard[2][0] = "a";
    keyboard[2][1] = "s";
    keyboard[2][2] = "d";
    keyboard[2][3] = "f";
    keyboard[2][4] = "g";
    keyboard[2][5] = "h";
    keyboard[2][6] = "j";
    keyboard[2][7] = "k";
    keyboard[2][8] = "l";
    keyboard[2][9] = "ò";
    // third row
    keyboard[3][0] = "z";
    keyboard[3][1] = "x";
    keyboard[3][2] = "c";
    keyboard[3][3] = "v";
    keyboard[3][4] = "b";
    keyboard[3][5] = "n";
    keyboard[3][6] = "m";
    keyboard[3][7] = ",";
    keyboard[3][8] = ".";
    keyboard[3][9] = "-";

    printMatrix(keyboard, x, y);

    List<Coordinata> cList = new ArrayList<Coordinata>();

    for (int i = 0, n = password.length(); i < n; i++) {
        char s = password.toLowerCase().charAt(i);
        cList.add(getCoordinate(s, keyboard));
    }
    int distanza = 0;

    for (int i = 0; i < cList.size()-1; i++) {
        distanza = distanza + distanceElement(cList.get(i), cList.get(i + 1));
    }

    System.out.println("distanza : " + distanza);
}


private static Coordinata getCoordinate(char s, String[][] keyboard) {
    Coordinata c = null;
    for (int i = 0; i < keyboard.length; ++i) {
        for (int j = 0; j < keyboard[0].length; ++j) {
            if (keyboard[i][j].equals(Character.toString(s))) {
                // Found the correct i,j 
                return c = new Coordinata(i, j, Character.toString(s));
            }
        }
    }
    return c;
}

private static int distanceElement(Coordinata c1, Coordinata c2) {
    int distance = 0;       
    distance = Math.abs(c2.getX() - c1.getX()) + Math.abs(c2.getY() - c1.getY());
    System.out.println("Distance: " + c1.getLettera() + "->" + c2.getLettera() + " = " + distance);     
    return distance;
}

private static void printMatrix(String[][] matrix, int matrixRow, int matrixCol) {
    System.out.println("Matrix is : ");
    for (int i = 0; i < matrixRow; i++) {
        for (int j = 0; j < matrixCol; j++) {
            System.out.print(matrix[i][j] + "\t");
        }
        System.out.println();
    }
}

检查QWERTY键盘中连续字母的算法

2 个答案: