java - 如何将两个字符串与随机位置android结合起来

Question

我正在尝试生成一个许可证 key ，其中两个字符串组合成另一个字符串。

String key1 = "X1X2X3X4..Xn"; //this is for the imei key cellphone
String key2 = "Y1Y2Y3Y4M1M2D1D2H1H2m1m2"; //this is a key to mix with the first one

组合的结果应该是这样的:

String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";

我像这样每两个空格拆分我的字符串，然后保存到一个数组中:

String [] key1splited = splitStringEvery(key1, 2);
String [] key2splited = splitStringEvery(key2, 2);

public String[] splitStringEvery(String s, int interval) {
    int arrayLength = (int) Math.ceil(((s.length() / (double)interval);
    String[] result = new String[arrayLength];

    int j = 0;
    int lastIndex = result.length - 1;
    for (int i = 0; i < lastIndex; i++) {
        result[i] = s.substring(j, j + interval);
        j += interval;
    } 
    result[lastIndex] = s.substring(j);

    return result;
}

我怎样才能使我的字符串组合得到如下所示的结果:

String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";

我希望有人能告诉我如何解决这个问题。

我正在尝试做这样的事情，但这是非常糟糕的方法:

static String res = "";
String[] key1splited = splitStringEvery(key1, 2);
String[] key2splited = splitStringEvery(key2, 2);
for (int i = 0; i < key2splited.length; i++) {
    if (key2splited[i].equals("D1")) {
        res = key2splited[i];
    }
    if (key2splited[i].equals("H1")) {
        res += key2splited[i];
        for (int j = 0; j < key1splited.length; j++) {
            if (key1splited[j].equals("X1")) {
                res += key1splited[j];
            }
            if (key1splited[j].equals("X2")) {
                res += key1splited[j];
            }
            if (key1splited[j].equals("X3")) {
                res += key1splited[j];
            }
        }
    }
}

等等，但这不是一个好的方法，因为字符串会发生变化。

Answer 1

我认为最简单的方法是将 token 从每个键中分离出来(2 个字符的长 token ，如 X1、D1 等)，然后将这些 token 组合成一个 String .

此外，您打乱 String，提取随机标记并构建许可证 key :

Random rand = new Random(); // generate Random object only once for efficiency purposes

// ... rest of your code

String getLicenceKey(String key1, String key2){
    List<String> tokens = new ArrayList<String>();

    // add tokens from key1
    for(int i = 0; i < key1.length(); i += 2) {
        tokens.add(key1.substring(i, i + 2));
    }

    // add tokens from key2
    for(int i = 0; i < key2.length(); i += 2) {
        tokens.add(key2.substring(i, i + 2));
    }

    // build the random result out of the tokens
    StringBuilder result = new StringBuilder();
    while(tokens.size() != 0){
        int randomPos = rand.nextInt(tokens.size());
        result.append(tokens.remove(randomPos));
    }

    return result.toString();
}

您提供的输入的示例输出:

m2XnD2m1H2X1..Y3Y1Y2Y4M1X2M2D1H1X4X3
H2Y2X4M1M2H1Y3Y1m2X1X2D1m1Xn..X3Y4D2
X1X4X3H2D2H1..M2m2Y3m1Y4M1D1Y1X2XnY2

key1="A1B1C1", key2="D1E1F1"的示例输出:

D1F1B1A1C1E1
C1A1D1B1E1F1
F1E1B1C1D1A1