我正在尝试使用两个字符串组合成另一个字符串来生成许可证密钥。
String key1 = "X1X2X3X4..Xn"; //this is for the imei key cellphone
String key2 = "Y1Y2Y3Y4M1M2D1D2H1H2m1m2"; //this is a key to mix with the first one
组合的结果应如下所示:
String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";
我像这样每隔两个空格分割字符串,然后保存到数组中:
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;
}
我怎样才能使我的琴弦组合得到如下所示的结果:
String result = "D1H1X1X2X3Y2Y4X4X5...XnY3Y1D2m2m1H1H2";
我希望有人能给我一个解决方法。
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];
}
}
}
}
依此类推,但这不是一个好方法,因为字符串将要更改。
我认为最简单的方法是将令牌与每个密钥分开(2个字符长的令牌,如X1,D1等),然后将令牌合并为一个String
。
此外,您还可以重新整理String
,提取随机令牌并构建许可证密钥:
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();
}
您提供的输入的样本输出:
m2XnD2m1H2X1..Y3Y1Y2Y4M1X2M2D1H1X4X3
H2Y2X4M1M2H1Y3Y1m2X1X2D1m1Xn..X3Y4D2
X1X4X3H2D2H1..M2m2Y3m1Y4M1D1Y1X2XnY2
key1 =“ A1B1C1 ”,key2 =“ D1E1F1 ”的示例输出:
D1F1B1A1C1E1
C1A1D1B1E1F1
F1E1B1C1D1A1
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句