在Java中生成唯一随机数的更好方法

Question

I need unique random ints in specified range. I use this approch:

class Main
{
static final int RANGE = 100;

static int uniqueGenerator(int range_, boolean boolArr_[], Random rand_)
{
    int tmpVar = rand_.nextInt(range_);
    while (boolArr_[tmpVar] == true)
    {
        tmpVar = rand.nextInt(range_);
    }
    boolArr_[tmpVar] = true;
    return tmpVar;
}

public static void main(String[] args)
{
    Random rand = new Random();  
    boolean boolArr[] = new boolean[RANGE];
    Arrays.fill(boolArr, false);
    int ceiling = 10;
    int tmp = Main.uniqueGenerator(ceiling, boolArr, rand);
    System.out.println(tmp); => 5
    ceiling = 20;
tmp = Main.uniqueGenerator(ceiling, boolArr);
    System.out.println(tmp); => 17

}
}

It seems to be cumbersome. Maybe someone knows better approach? EDIT: I use it in game code, so I need most efficient solution. Answers below suggest initializing new list, shuffling it => too resource consuming/need to generate new list every time when need to change range.

Answer 1

List<Integer> list = new ArrayList<Integer>();
for(int i = 1; i <= 100; i++) list.add(i);
Collections.shuffle(list);

Answer 2

Fill an array with the range of numbers you want, shuffle it and extract an item.

Edit: Look at Eng.Fouad's example to see how this is implemented.

Answer 3

Generate and store random numbers in set

Set<Integer> set = new HashSet<Integer>(100);
    Random rand = new Random();
    while (set.size() < 1000) {
        set.add(rand.nextInt(100));
    }

    for (Integer integer : set) {
        System.out.println(integer);
    }