Java Array of unique randomly generated integers

Question

public static int[] uniqueRandomElements (int size) {

    int[] a = new int[size];

    for (int i = 0; i < size; i++) {
        a[i] = (int)(Math.random()*10);
         


        

Answer 1

for (int i = 0; i < size; i++) {
    a[i] = (int)(Math.random()*10);

    for (int j = 0; j < i; j++) {
        if (a[i] == a[j]) {
            a[j] = (int)(Math.random()*10); //What's this! Another random number!
        }
    }   
}

You do find the duplicate values. However, you replace it with another random number that may be a duplicate. Instead, try this:

for (int i = 0; i < size; i++) {
    a[i] = (int)(Math.random()*10);//note, this generates numbers from [0,9]

    for (int j = 0; j < i; j++) {
        if (a[i] == a[j]) {
            i--; //if a[i] is a duplicate of a[j], then run the outer loop on i again
            break;
        }
    }  
}

However, this method is inefficient. I recommend making a list of numbers, then randomizing it:

ArrayList a = new ArrayList<>(11);
for (int i = 0; i <= 10; i++){ //to generate from 0-10 inclusive. 
                               //For 0-9 inclusive, remove the = on the <=
    a.add(i);
}
Collections.shuffle(a);
a = a.sublist(0,4);
//turn into array

Or you could do this:

ArrayList list = new ArrayList<>(11);
for (int i = 0; i <= 10; i++){
    list.add(i);
}
int[] a = new int[size];
for (int count = 0; count < size; count++){
    a[count] = list.remove((int)(Math.random() * list.size()));
}