如何生成特定范围内的唯一随机数

Question

i want to generate 255 unique random numbers within this range (0-255). so that the array will not contain duplicate records

short [] array =new short[255];
Random rand = new Random();
boolean   flag=false;
for (int i=0;i<array.length;i++){
    int random_integer = rand.nextInt(255-0) + 0;
    for (int j=0;j<i;j++){
        if ((short)random_integer==array[j]){
            flag=true;
        }
    }
    if (flag==false){
        array[i]=(short)random_integer;  
    }
}
for (int i=0;i<array.length;i++){
    System.out.println(array[i]);
} 

but i get only first 20 0r 30 items with values and the rest of the array items equals zero.

Answer 1

Solution 1:

I read Jon Skeet's comment, of course, this is the easiest solution:

List<Integer> list = new ArrayList<>();
for (int i = 0; i < 255; i++) {
     list.add(i);
}
//and here is the point. Java already have this implemented for you
Collections.shuffle(list);

Or in Java 8 declarative style:

List<Integer> list= IntStream.range(0, 255)
    .boxed()
    .collect(Collectors.toList());
Collections.shuffle(list);

or

List<Integer> list = new ArrayList<>();
IntStream.range(0, 255).forEach(list::add);
Collections.shuffle(list);

Solution 2 (picking up on your solution):

You need to generate number for each cell, and check if this number already exists:

 short [] array =new short[255];
 Random rand = new Random();

 for (int i=0; i<array.length; i++) {
     int random_integer = -1;

     //generate integer while it exists in the array
     while(exists(random_integer, array)) {
         random_integer = rand.nextInt(255);
     }

     array[i] = random_integer;
}

And now, let's check whether it exists:

public boolean exists(int number, int[] array) {
    if (number == -1)
        return true; 

    for (int i=0; i<array.length; i++) {
        if (number == array[i])
            return true;
    }
    return false;
}

Of course, you can use a hashmap for example, to speed up the exists() method, ie to lower the complexity from O(n) to O(1);

Answer 2

如果可以使用Java 8：

List<Integer> randIntegers = new Random().ints(1, 256).distinct().limit(255).boxed().collect(Collectors.toList());

Answer 3

You check if the random number existed, but you do not reset the flag to false in your loop, so as soon as the first duplicate number comes, nothing happens anymore, because flag is always true.

You also should build in somthing like:

if ((short)random_integer==array[j]){
  flag=true;
  i--;
}

to make sure to revisit an index of your array after you skipped it for a duplicate number.

Answer 4

public static void main(String ar[]){
short [] array =new short[255];
Random rand = new Random();
int random_integer;
boolean   flag=false;
for (int i=0;i<array.length;i++){
     random_integer = rand.nextInt();
     for (int j=0;j<i;j++){
         if ((short)random_integer==array[j]){
                 flag=true;
                 i--;
            }
      }
      if (flag==false)
        array[i]=(short)random_integer;  
}
for (int i=0;i<array.length;i++)
    System.out.print(" "+array[i]);
System.out.println();
}

Answer 5

You need to reset flag value in every iteration or change your logic:

       for (int j=0;j<i;j++){
                flag=false;//<--
                if ((short)random_integer==array[j]){
                    flag=true;
                }
            }

Answer 6

Did you considered creating an HashSet and putting values you've used in it? Then your code should look like this:

HashSet hs = new HashSet();
short [] array =new short[255];
Random rand = new Random();

for (int i=0;i<array.length;i++){
int random_integer = rand.nextInt(255-0);
if (!hs.contains(random_integer ))
{
array[i]=(short)random_integer;  
hs.put(random_integer);
}
else{ //generate new integer}
}

Answer 7

It is very hard to read the code without the proper indentation.

Anyhow - you don't do anything if flag == true . So obviously you don't fill many places in the array.