Question

public class WordScrambleEx1 {
    public static void main(String[] args) {
        String[] strArr = {"CHANGE", "LOVE", "HOPE", "VIEW"};

        String answer = getAnswer(strArr);
        String question = getScrambledWord(answer);

        System.out.println("Question :" + question);
        System.out.println("Answer: " + answer);
    }


    public static String getAnswer(String[] strArr) {

        String i = strArr[(int)Math.random()*4]; 
        return i;


    }
    public static String getScrambledWord(String str) { 

        char[] character = str.toCharArray();
        String question1 = null;
        for(int i = 0; i < character.length; i ++)
        {
        char[] java = new char [(int)Math.random()*i] ;
        question1 = new String(java);
        }

        return question1;
    }   
}

我对Java很陌生并且给出了一个问题，我给出了四个字母，我的方法需要使用Math.random随机选择其中一个并加扰该字符串的字符。

我的代码从给定数组中找到一个String，但不对字符串进行加扰。谁能告诉我我做错了什么？

理解构造函数和范围真的很难。

Answer 1

第一个错误：

(int) Math.random() * i

将始终返回0，因为Math.random（）返回一个介于0和1之间的浮点数，因此当它转换为int时它总是为零（int不圆，它只是截断逗号后面的数字）。

您可以使用以下方法解决此问题：

(int) (Math.random() * i)

现在我们首先将Math.random（）的float结果与i相乘，得到一个float，因为第一个数字是float。然后我们将这个float转换为int。

第二个错误：

public static String getScrambledWord(String str) { 

char[] character = str.toCharArray();
String question1 = null;
for(int i = 0; i < character.length; i ++)
{
char[] java = new char [(int)Math.random()*i] ;
question1 = new String(java);
}

return question1;
}

每次迭代都会创建一个长度为0的新char数组，然后将question1设置为它，它总是一个空字符串，因为java数组中没有任何内容。

我会这样做：

public static String getScrambledWord(String str) {
    char[] character = str.toCharArray();
    String question1 = new String();

    ArrayList<Character> chars = new ArrayList<Character>(); //an arraylist is an array wich dynamically changes its size depending on the amount of its elements
    for (int i = 0; i < character.length; i++) {// first we put all characters of the word into that arraylist
        chars.add(character[i]);
    }

    while(chars.size()>0){//then we iterate over the arraylist as long as it has more than 0 elements
        int index = (int)(Math.random() * chars.size());//we create a random index in the range of 0 and the arraylists size
        question1 += chars.get(index);// we add the letter at the index we generated to the scrambled word variable
        chars.remove(index);// then we remove the character we just added to the scrambled word, from the arraylist, so it cant be in there twice
    }// thus the size decreases by 1 each iteration until every element of the arrraylist is somewhere in the scrambled word

    return question1;
}

Answer 2

您的代码中存在一些错误。生成随机整数的方式具有误导性。让我们看一下(int)Math.random() * 4语句的解释。 Math.random()确实：

返回带有正号的double值，大于或等于0.0且小于1.0。

现在，在Java中，类型转换优先于+，-，*和/，因此实际发生的是((int)Math.random()) * 4。 Math.random()会在0.0和1.0 独占之间返回一个浮点数，因此大致为[0.0,0.999999 ...]。对int的强制转换将截断所有小数位，您将始终获得0。然后，您的陈述将简化为0 * 4 = 0。总的来说，你总是得到第一个字。

我建议你改用Random课程。它提供了一种方法nextInt(int n)，它返回0 包含和n 独占之间的随机整数，因此[0, n - 1]

由于代码中存在很多错误，我想为您提供此解决方案：

import java.util.Random;

public class WordScrambleEx1  {

    private static Random random;

    public static void main(String[] args) {

        // Create object of class (initializes the
        // random generator with a default seed)
        random = new Random();

        String[] strArr = { "CHANGE", "LOVE", "HOPE", "VIEW" };

        String answer = getAnswer(strArr);
        String question = getScrambledWord(answer);

        System.out.println("Question: " + question);
        System.out.println("Answer: " + answer);

    }

    public static String getAnswer(String[] strArr) {

        // Chooses a random index in [0, strArr.length - 1]
        int index = random.nextInt(strArr.length);

        String i = strArr[index];
        return i;

    }

    public static String getScrambledWord(String str) {

        String remaining = str;
        String scrambled = "";

        // Loop over the string, each time choose a random letter
        // and add it to the scrambled word, then remove that letter
        // from the remaining word. Repeat until all letters are gone.
        for (int i = str.length(); i > 0; i--) {

            // Choose the index of a letter in the remaining string
            int index = random.nextInt(remaining.length());

            // Add the letter at the random index to your scambled word
            scrambled += remaining.charAt(index);

            // Remove the chosen character from the remaining sequence
            remaining = remaining.substring(0, index) + remaining.substring(index + 1);

        }

        return scrambled;
    }

}

字符串的加扰字符

2 个答案: