Question

所以我试着学习Java，我编写了这段代码，假设从列表中生成单词组合，然后放在一个句子中。问题是第一个列表中随机选择的单词（名称）只包含名称，将被重用（我知道为什么，但我不确定我是否需要“phrase3”或第三个列表。

这是我的代码：

    package ListOfWords;

     public class test {
   public static void main (String[] args) {
   String[] name = {"Nicole","Ronnie","Robbie","Alex","Deb"};
   String[] action = {"liar","driver","cook","speller","sleeper","cleaner","soccer   
         player"};

 // find out how many words there are in each list
   int nameLength = name.length;
   int actionLength = action.length;

 // Generate two random numbers 
   int  rand1 = (int) (Math.random() * nameLength);
   int  rand2 = (int) (Math.random() * actionLength);

   String phrase1 = name[rand1];
   String phrase2 = action[rand2];

   System.out.print("It is obvious that" + ' ' + phrase1 + " " + "is a better" + " " +  
   phrase2 + " " + "than" + " " + phrase1 + "!" );          
   }
 }

这是我现在得到的结果：

    It is obvious that Robbie is a better cleaner than Robbie!

因此，当您看到第一个列表中的随机名称被重用时 - 我如何确保它不会从第一个列表中选择相同的元素（名称）？

Answer 1

您需要第三个随机数和短语，以选择要使用的第二个名称。例如：

// Generate two random numbers 
   int  rand1 = (int) (Math.random() * nameLength);
   int  rand2 = (int) (Math.random() * actionLength);
   int  rand3 = (int) (Math.random() * nameLength);

   String phrase1 = name[rand1];
   String phrase2 = action[rand2];
   String phrase3 = name[rand3];

   System.out.print("It is obvious that" + ' ' + phrase1 + " " + "is a better" + " " +  
   phrase2 + " " + "than" + " " + phrase3 + "!" );

编辑：为避免为phrase1和phrase3选择相同的名称，以下代码应确保使用不同的索引来选择phrase3而不是phrase1：

int  rand1 = (int) (Math.random() * nameLength);
int  rand2 = (int) (Math.random() * actionLength);
int  rand3 = (int) (Math.random() * nameLength);
while(rand1==rand3){
    rand3 = (int) (Math.random() * nameLength);
}

这将导致rand3被更改，直到它与rand1不同，后者将为phrase1和phrase3选择不同的名称。

请注意，如果names数组中只有一个名称，则会导致无限循环。

Answer 2

你可以这样做：

List<String> randomNames = new ArrayList(Arrays.asList(name));
Collections.shuffle(randomNames);

int randAction = (int) (Math.random() * actionLength);

String phrase1 = randomNames.get(0);
String phrase2 = action[randAction];
String phrase3 = randomNames.get(1);

System.out.print("It is obvious that " +  phrase1 + " is a better " 
     +  phrase2 + " than " + phrase3 + "!" );

Answer 3

   //Generate two random numbers 
   int  rand1 = (int) (Math.random() * nameLength);
   int  rand2 = (int) (Math.random() * actionLength);
   int  rand3;
   do{
       rand3 = (int) (Math.random() * nameLength)
   } while (rand3 == rand1);

   String phrase1 = name[rand1];
   String phrase2 = action[rand2];
   String phrase3 = name[rand3];

   System.out.print("It is obvious that" + ' ' + phrase1 + " " + "is a better" + " " +  
   phrase2 + " " + "than" + " " + phrase3 + "!" );

Answer 4

例如，您似乎将随机初始化为4。然后，每次调用该索引时，您获得的值都相同。

在该结构中，您需要一个新变量。

如果你看一下编程的流程，你可以创建两个radom。然后设置它们。他们决不会重新初始化。

添加另一个变量来解决或创建一个函数来返回一个新的rand并将其传递给我而不是

看到你对第3阶段的评论可能是相同的。

来自下面的评论。首先创建一个包含名称列表索引的数组。随机选择一个值。将此索引值替换为列表中的最后一个，并选择长度为1的另一个值。 - Jongware 27分钟前。魔法。

句子生成器与列表中的单词

4 个答案: