Question

I m试图生产一个许可证钥匙,有两条护法将两者合并成另一条str。

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

组合的结果应如此:

String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";

我将座右铭分成两处,我只把座椅分成一阵:

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;
}

我怎么能够把我的发言结合起来,给我带来这样的结果:

String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";

我希望有人能给我一个如何解决这一问题的想法。

我正试图做这样的事情,但这种方法非常糟糕:

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];
            }
        }
    }
}

因此,但这不能成为这样做的良好途径,因为扼杀物会发生变化。

Answer 1

我认为,你最容易做到的办法是将脚印从每个钥匙中分离出来(如X1、D1等2个长期标注),然后把标语合并成一个<条码>。

此外,你填写了<代码>St,抽取随机标本,并建造了许可证钥匙:

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();
}

你提供的投入的样本:

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

关键1=“A1B1C1>、关键2=<>D1E1F1:

D1F1B1A1C1E1
C1A1D1B1E1F1
F1E1B1C1D1A1

Answer 2

I m hurry so if i get it, you want something like this? This may give you some ideas using ArrayList. If someone find something wrong fell free to edit.

    String tmp = "X1X2X3";
    String tmp2 = "Y1Y2Y3";

    ArrayList<String> list = new ArrayList<String>();

    for(int i = 0 ; i < (tmp.length()/2) ; i  = i + 2)
    {
        list.add(tmp.substring(i,i+1));
        list.add(tmp2.substring(i,i+2));
    }

    int max = list.size();
    String finalMix = null;

    for(int j = 0 ; j < max; j++)
    {
        Random rnd = new Random();
        int rndNumber = rnd.nextInt((list.size()) + 1);
        finalMix += list.get(rndNumber);
        list.remove(rndNumber);
    }
    System.out.println(finalMix);
}

Answer 3

查阅这一SO问题:。我如何在 Java的具体范围内产生随机分类的分类?

随机抽出数字,并用你在座标的“部分”中的随机指数进行重建。

Answer 4

我建议采用简单的加密法,如:

org.apache.commons.codec.digest.DigestUtils.sha384Hex(key1+key2);

This yields always the same result for the same key1,key2 pair, but cannot be used to calculate any of them.
You can truncate the resulting 24 bytes, but of course you somewhat increase the risk of collisions (you re still pretty good with 12 bytes).
You need a different strategy if you need to reverse the process (i.e. get key1,key2 from the result), or simply store those two server-side with the result if you need to (this may create privacy issues).

Answer 5

我最近的面谈逻辑之一

public class Main {
  public static void main(String[] args) {
  
  String a = "abcd";

  String b = "123456";

  StringBuffer sb = new StringBuffer(b);
  
  for(int i=0;i<4;i++){

  sb.insert(i*2,a.charAt(i));

 }
    
    System.out.println(sb);
  }
}

友情链接