how to generate the array contains random number in ascending order using java

Question

I would like to make the array contains random number in ascending order, for example (1,5,10...100), but not (5,1,10...). this code but when run the number not be ascending. what the error?

     Random r=new Random();
     int t=10;
       int a[]=new int[t];
     int count=0;
     int end=0;
     int curr=0;
     while(count<t){
     curr=r.nextInt();
     end=end+curr;
     count++;}
     for(int i=0;i<to;i++){
     a[i]=r.nextInt(10);}
 ```

Answer 1

You can use the Random#ints method which returns a stream of random int values. Assuming you want 10 random ints from the range [1,100) sorted in ascending order:

Random r = new Random();
int t    = 10;
int a[]  = r.ints(1, 100).distinct().limit(t).sorted().toArray();
System.out.println(Arrays.toString(a));

you can omit distinct if you want to allow duplicates

To sort in descending order you need to box the primitiv types to Integer and use Comparator.reverseOrder() or Collections.reverseOrder() and unbox them after sorting to store them in an int array

int a[]  = r.ints(0, 100)
                .distinct()
                .limit(t)
                .boxed()
                .sorted(Comparator.reverseOrder())
                .mapToInt(Integer::intValue)
                .toArray();