在随机位置替换字符串中的随机字符数

Question

我正在尝试编写一个程序来读取一个包含 4000 或 (x) 个单词的 txt 文件。到目前为止，我设法编写了一个程序，如下所示。它有点像将（随机）星数（*）添加到单词中的随机位置。然而，问题就在这里。它总是添加一个 *。我该如何更改它，以便它可以为单词添加随机数量的星星。

例如：让我们以母亲这个词为例。我想要一个程序也将随机数 * 添加到此 word.That 中的随机位置，在此示例中，随机数的星星将是 0、1、2、3、4、5 或 6。如果我运行多次编程，我会得到不同的结果，例如：（* 在下面显示为 ?）

M??her , Mot??r, Mother, ?other, ??????, M????r ,ect.

欢迎提出任何想法。

public class random_menjava_z_zvezdico {

public static void main(String[] args) {
    String test;
    int dolzina = 0;
    String outputFile = "random_z_zvezdico.txt";

    ArrayList<String> list = new ArrayList();


    try {

        File file = new File("4000_random_besed.txt");
        FileReader fileReader = new FileReader(file);
        BufferedReader bufferedReader = new BufferedReader(fileReader);
        String vrstica = bufferedReader.readLine();
        while (vrstica != null) {

            list.add(vrstica);
            vrstica = bufferedReader.readLine();
            // dolzina=list.size();
            // System.out.println(dolzina);

        }

        FileWriter fileWriter = new FileWriter(outputFile);
        PrintWriter out = new PrintWriter(fileWriter);
        System.out.println("4000 besedam je bila dodana * random krat na random mestu!");
        for (int idx = 0; idx < list.size(); ++idx) {
            test = list.get(idx);
            dolzina = test.length();

            Random rGenerator = new Random();
            StringBuilder beseda = new StringBuilder(test);

            for (int i = 0; i < dolzina; ++i) {
                int randomInt = rGenerator.nextInt(dolzina);
                beseda.setCharAt(randomInt, '*');

            }
            out.println(beseda);

        }out.close();
    } catch (IOException e) {
        e.printStackTrace();

    }
}
}

Answer 1

我会改变：

for (int i = 0; i < dolzina; ++i) {
    int randomInt = rGenerator.nextInt(dolzina);
    beseda.setCharAt(randomInt, '*');
}

对于：

for (int i = 0; i < dolzina; i++) {
    int randomInt = rGenerator.nextInt(2);
    if (randomInt > 0) {
        beseda.setCharAt(i, '*');
    }
}

但更好的解决方案是：

StringBuilder orig = new StringBuilder("Mother");
Random rnd = new Random();

int maskSize = rnd.nextInt(orig.length()+1);

for (int i = 0; i < maskSize; i++) {
    int pos = rnd.nextInt(orig.length());

    while (orig.charAt(pos) == '*') {
        pos++;
        if (pos >= orig.length()) {
            pos = 0;
        }
    }

    orig.setCharAt(pos, '*');
}

Answer 2

int randomInt = rGenerator.nextInt(7);//this gets you a number between 0 and 6 (inclusive)

然后如果你想更改随机字符数

int positionCount = 0;//this wiil be used to change the character at specific location starting from the first location

while(randomInt >= 0){
//change this line to randomInt > 0 if you don't want to change anything when you have zero as randomInt

  beseda.setCharAt(positionCount, '*');//this replace the next character starting from the first

  //control the values in order for the position not to exceed the string by using if statement
  if(positionCount < beseda.toString().length){
     positionCount++;
  }

  randomInt--;//decrease the random number by one to enable escape of while loop when it is done
}

或者您也可以使用之前生成的相同随机整数更改随机位置的字符：

beseda.setCharAt(randomInt, '*');

Answer 3

我看到您没有定义不同配置所需的概率分布。我在我的解决方案中假设你需要一个统一的概率（即 "mother"/"*other"/"**ther" 的概率都是 1/64）：

  private static final Random rnd = new Random();

  public static String replaceRandom(String s) {
    char[] chars = s.toCharArray();
    BitSet bs = new BitSet();
    byte[] bytes = new byte[chars.length/8 + 1];
    rnd.nextBytes(bytes);
    bs = BitSet.valueOf(bytes);
    int i = bs.nextSetBit(0);
    while (i < chars.length && (i != -1)) {
      chars[i] = '*';
      i = bs.nextSetBit(i + 1);
    }

    return new String(chars);
  }

在随机位置替换字符串中的随机字符数

Replacing random number of characters in a string on random places

java

random