fill an array with integers 1-10 while checking to see if the integer is already in the array

Question

I am filling an array with random integers 1 - 10, however while I'm doing this I need to check to see if the Integer is already in the array. I have no problem filling the array with the integers but my code to check for duplicate integers is not working properly.

  package arrayPackage;
  import java.util.Arrays;
  import java.util.Random;
  public class BruteForce 
  {
//declare an integer array with 10 numbers.
int[] array = new int[10];
Random randomInt = new Random();    //random number generator


public void shuffle()
{ 
    //int prevInt=0;
    for(int i = 0; i < array.length; i++)   //for each integer in the array pick a random integer and place in array
    {
        int temp = 1 + randomInt.nextInt(array.length);
        for(int j = i; j < array.length; j++)
        {
            if(temp != array[j] && temp!= array[i])
            {
                array[i] = temp; 
            }
        }

    }
}

public void displayArray()
{
    System.out.println(Arrays.toString(array));
}

}

Wouldn't it be easier to fill the array with numbers 1-10, then shuffle it? — Kayaman, Oct 23 '14 at 06:52
You fill in the array with `temp` when there is no duplicate, but where do you handle if there is duplicate? — Pham Trung, Oct 23 '14 at 06:55
implement the [fisher-yates-knuth shuffle](https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle) it is efficient and unbiased, else use a ready-made implementation (built-in method in java) — Nikos M., Aug 21 '15 at 11:33

score 1 · Answer 1 · answered Oct 23 '14 at 07:17

1

seems that you want an array with 1,2,3,...,10 random distributed.

The following solution is better since it only needs exactly 10 random calls:

int array = new int[10];

for (int i = 0; i < 10; i++)
{
  array[i] = i + 1;
}

for (int i = 0; i < 10; i++)
{
  int to = randomInt.nextInt(array.length - i) + i; // get random swap target between i and 9
  swap(array[i], array[to]); // I assume Java has swap function
}

answered Oct 23 '14 at 07:17

Van Yu

129
1
11

Java doesn't have a `swap`-method, so you have to make a simple one yourself: `private void swap(int[] array, int i, int to){ int temp = array[to]; array[to] = array[i]; array[i] = temp; }` or with some checks: `private void swap(int[] array, int i, int to){ if(array != null && array.length > 2 && i < array.length && to < array.length && i != to){ int temp = array[to]; array[to] = array[i]; array[i] = temp; } }` – Kevin Cruijssen Oct 23 '14 at 07:36

score 1 · Answer 2 · edited May 23 '17 at 11:51

If your goal is to have an int-array with the numbers 1 to 10, and then shuffle them, try this:

private static int[] array = new int[10];

public static void main(String[] args){     
    // Fill the array
    for(int i = 0; i < array.length; i++)
        array[i] = i + 1;

    shuffle(array);
    System.out.println(Arrays.toString(array));
}

private static void shuffle(int[] array){
    if(array != null && array.length > 2){
        Random randomGenerator = new Random();
        for(int i = array.length - 1; i > 0; i--){
            int randomIndex = randomGenerator.nextInt(i + 1);
            // swap
            int a = array[randomIndex];
            array[randomIndex] = array[i];
            array[i] = a;
        }
    }
}

If your goal is to have an int-array and every time you want to add a random int check wether or not it doesn't exist yet, try using this algorithm to get a random int with excludes:

public int getRandomWithExclusion(Random rnd, int start, int end, int... exclude) {
    int random = start + rnd.nextInt(end - start + 1 - exclude.length);
    for (int ex : exclude) {
        if (random < ex) 
            break;

        random++;
    }
    return random;
}

For information on how to use it, see this SO answer here. The only issue is with the excludes-array, which should change size every time so it only contains the excluded ints, instead of also the default 0 when you create an int-array of size 10. Otherwise exclude.length always returns 10, making the algorithm useless.

and if one wants to shuffle only part of an array (either with included or excluded items) one can use [a variation of fisher-yates-knuth shuffle](http://stackoverflow.com/questions/26718843/php-shuffle-only-part-of-an-array/28491007#28491007) to do it efficiently — Nikos M., Aug 21 '15 at 11:37

score 1 · Answer 3 · answered Oct 23 '14 at 08:14

1

It is very easy to write a flawed shuffle algorithm, that is why there exists a good shuffler built in:

http://docs.oracle.com/javase/6/docs/api/java/util/Collections.html#shuffle(java.util.List)

answered Oct 23 '14 at 08:14

weston

54,145
21
145
203

Gianluca Colombo · Answer 4 · 2014-10-23T07:29:54.853

I've found this that is probably the solution you are searching for..

function shuffle(array) {
  var copy = [], n = array.length, i;

  // While there remain elements to shuffle…
  while (n) {

    // Pick a remaining element…
    i = Math.floor(Math.random() * array.length);

    // If not already shuffled, move it to the new array.
    if (i in array) {
      copy.push(array[i]);
      delete array[i];
      n--;
    }
  }

  return copy;
}

Usage:

public void displayArray()
{
  var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
  shuffle(arr);
  System.out.println(Arrays.toString(arr));
}

score 0 · Answer 5 · answered Oct 23 '14 at 07:38

0

Why dont you use HashSet<Integer> if you dont want duplicate elements in the list.

Later you convert them to array on your wish.

answered Oct 23 '14 at 07:38

Rookie007

1,229
2
18
50

How about `LinkedHashSet` ? – Rookie007 Oct 24 '14 at 01:09

fill an array with integers 1-10 while checking to see if the integer is already in the array

5 Answers5