How to select random values from a given range

问题:

I'm trying to create an android application which will generate random series of values (integer numbers in this case) in a given range (but NOT equal between them) and display them in a simple TextView

Let's say we have the range R = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

Each time I press the button "Generate" I want to randomly generate 5 different results

Example for each "Generate":

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

EDIT (works now) Thanks for all the help!

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);
  }
}

回答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]);
}

data contains your range numbers.

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

set this text in your TextView.

回答2:

in the edit code:

int idx = r.nextInt();

needs to change to:

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

because you want to choose a random index from your data.

回答3:

If I understand this correctly, you all the numbers should be unique. You could fill a list with the range you want to draw from. Every time you have selected a value from it, you should remove it from the list so it won't be selected a second time. I hope this description is clear enough. If not, I will provide some code.

回答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)
};

回答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;
}

回答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;

hope use full to you....