Question

我正在尝试创建一个Android应用程序，它将在给定范围内生成随机序列值（在这种情况下为整数）（但 NOT 在它们之间相等）并在简单中显示它们TextView的

假设我们的范围 R = [1,2,3,4,5,6,7,8,9,10,11,12,13]

每次按下按钮 “生成” 我想随机生成5个不同的结果

每个“生成”的示例：

4,9,2,12,10
5,1,6,8,13
10,4,6,8,2
等...

编辑（现在有效）感谢您的帮助！

public class random extends Activity {


static final Integer[] data = new Integer[] {
    1, 2, 3, 4, 5, 6, 7, 8
    };


@Override
public void onCreate(Bundle icicle) {
    super.onCreate(icicle);
    setContentView(R.layout.main);

    Random r = new Random();

    Set<Integer> mySet = new HashSet<Integer>();
    while (mySet.size() < 5) {
       int idx = r.nextInt(data.length);
       mySet.add(data[idx]);
    }

    String text = "";
    for (Integer i : mySet) {
       text = text + i + ", "; 
    }
       TextView Numbers = (TextView)findViewById(R.id.shownumbers);
       Numbers.setText(text);
  }
}

Answer 1

Random r = new Random(<a seed number>);

Set<Integer> mySet = new HashSet<Integer>();
while (mySet.size() < 5) {
   int idx = r.nextInt(<length of your data>)
   mySet.add(data[idx]);
}

数据包含您的范围编号。

String text = "";
for (Integer i : mySet) {
   text = text + i + ", "; 
}

在TextView中设置此文本。

Answer 2

在编辑代码中：

int idx = r.nextInt();

需要更改为：

int idx = r.nextInt(data.length);

因为您想从数据中选择一个随机索引。

Answer 3

如果我理解正确，你所有的数字都应该是唯一的。您可以使用要绘制的范围填充列表。每次从中选择一个值时，都应该从列表中删除它，这样就不会再次选择它。我希望这个描述足够清楚。如果没有，我会提供一些代码。

Answer 4

int low = ...;
int high = ...;
List<Integer> choices = new LinkedList<Integer>();
for (int i = low; i <= high; i++) {
    choices.add(i);
} 

Collections.shuffle(choices);

int[] choices = new int[] {
  choices.get(0),
  choices.get(1),
  choices.get(2),
  choices.get(3),
  choices.get(4)
};

Answer 5

final StringBuilder builder = new StringBuilder();
final Random r = new Random(System.currentTimeMillis());
final Set<Integer> numbers = new HashSet<Integer>();

public String getRandomNumbers(int count,int min, int max)
{
    if(count > (max - min) || (max < min))
        throw new IllegalArgumentException("There is an error with the parameters provided");

    builder.setLength(0); //Clear StringBuilder
    numbers.clear(); //Clear the number list

    int i=0;
    while( i < count )
    {
        int aRandomNumber = (r.nextInt() % max) +min;

        if(numbers.contains(aRandomNumber)) // If we have seen this number already
            continue;
        else
        {
            i++;
            numbers.add(aRandomNumber);
            builder.append(aRandomNumber); //Add number to string

            if( i < (count-1) )
                builder.append(", "); // Add a comma if it's not the last number
        }
    }

    String output = builder.toString();
    builder.setLength(0); //A polite clearing
    numbers.clear();

    return output;
}

Answer 6

public int[] generateSeries(){
 int[] R = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};
    int [] newarray=new int[5];
Collections.shuffle(Arrays.asList(R));
for(int i=0;i<5;i++)
{
  newarray[i]=R[i];
  System.out.println("values => "+R[i]);
}
return newarray;

希望充分利用你......

如何从给定范围中选择随机值

6 个答案: