Generating Unique Random Numbers Within a Range

var arr = [];while(arr.length < 8){    var r = Math.floor(Math.random() * 100) + 1;    if(arr.indexOf(r) === -1) arr.push(r);}console.log(arr);

how to generate unique random numbers with a specific range

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, i.e. to lower the complexity from O(n) to O(1);

How do I generate random but unique numbers in python?

You can get such a list of a given number of non-repeating elements taken from a given pool via random.sample:

>>> random.sample(range(500, 510), 9)
[500, 501, 503, 502, 505, 507, 508, 506, 504]

How to generate unique random numbers (that don't repeat)?

One straightforward way to do non-repeating 'random' (psudeorandom) whole numbers in a modest range is to create a list using range(1, n), then random.shuffle() the list, and then take as many numbers as you want from the list using pop() or a slice.

import random

max = 11
l = list(range(1, max))  # the cast to list is optional in Python 2
random.shuffle(l)

Now every time you want a random number, just l.pop().

Another is to use random.sample() -- see https://docs.python.org/3/library/random.html

Generate N random and unique numbers within a range

Take an array of 50 elements: {1, 2, 3, .... 50}
Shuffle the array using any of the standard algorithms of randomly shuffling arrays. The first six elements of the modified array is what you are looking for. HTH

Generating UNIQUE Random Numbers within a range

Array with range of numbers at random order:

$numbers = range(1, 20);
shuffle($numbers);

Wrapped function:

function UniqueRandomNumbersWithinRange($min, $max, $quantity) {
    $numbers = range($min, $max);
    shuffle($numbers);
    return array_slice($numbers, 0, $quantity);
}

Example:

<?php
print_r( UniqueRandomNumbersWithinRange(0,25,5) );
?>

Result:

 Array
(
    [0] => 14
    [1] => 16
    [2] => 17
    [3] => 20
    [4] => 1
)

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