Math Random Number Without Repeating a Previous Number

math random number without repeating a previous number

There are a number of ways you could achieve this.

Solution A:
If the range of numbers isn't large (let's say less than 10), you could just keep track of the numbers you've already generated. Then if you generate a duplicate, discard it and generate another number.

Solution B:
Pre-generate the random numbers, store them into an array and then go through the array. You could accomplish this by taking the numbers 1,2,...,n and then shuffle them.

shuffle = function(o) {
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

var randorder = shuffle([0,1,2,3,4,5,6]);
var index = 0;

setInterval(function() {
    $('.foo:nth-of-type('+(randorder[index++])+')').fadeIn(300);
}, 300);

Solution C:
Keep track of the numbers available in an array. Randomly pick a number. Remove number from said array.

var randnums = [0,1,2,3,4,5,6];

setInterval(function() {
    var m = Math.floor(Math.random()*randnums.length);
    $('.foo:nth-of-type('+(randnums[m])+')').fadeIn(300);
    randnums = randnums.splice(m,1);
}, 300);

Generating non-repeating random numbers in JS

If I understand right then you're just looking for a permutation (i.e. the numbers randomised with no repeats) of the numbers 1-10?
Maybe try generating a randomised list of those numbers, once, at the start, and then just working your way through those?

This will calculate a random permutation of the numbers in nums:

var nums = [1,2,3,4,5,6,7,8,9,10],
    ranNums = [],
    i = nums.length,
    j = 0;

while (i--) {
    j = Math.floor(Math.random() * (i+1));
    ranNums.push(nums[j]);
    nums.splice(j,1);
}

So, for example, if you were looking for random numbers between 1 - 20 that were also even, then you could use:

nums = [2,4,6,8,10,12,14,16,18,20];

Then just read through ranNums in order to recall the random numbers.

This runs no risk of it taking increasingly longer to find unused numbers, as you were finding in your approach.

EDIT: After reading this and running a test on jsperf, it seems like a much better way of doing this is a Fisher–Yates Shuffle:

function shuffle(array) {
    var i = array.length,
        j = 0,
        temp;

    while (i--) {

        j = Math.floor(Math.random() * (i+1));

        // swap randomly chosen element with current element
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;

    }

    return array;
}

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

Basically, it's more efficient by avoiding the use of 'expensive' array operations.

BONUS EDIT: Another possibility is using generators (assuming you have support):

function* shuffle(array) {

    var i = array.length;

    while (i--) {
        yield array.splice(Math.floor(Math.random() * (i+1)), 1)[0];
    }

}

Then to use:

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

ranNums.next().value;    // first random number from array
ranNums.next().value;    // second random number from array
ranNums.next().value;    // etc.

where ranNums.next().value will eventually evaluate to undefined once you've run through all the elements in the shuffled array.

Overall this won't be as efficient as the Fisher–Yates Shuffle because you're still splice-ing an array. But the difference is that you're now doing that work only when you need it rather than doing it all upfront, so depending upon your use case, this might be better.

Random Numbers without repeat. Beginner Lvl

You can create an array of the numbers that you want to choose from. Then simply remove the number when it is pulled.

function lottoGame(max_numbers) {
  let lottoArray = [];
  let numbers = [];
  let max_num = 6;
  for (var i = 1; i <= max_num; i++) {
    numbers.push(i);
  }
  nums = [];
  for (let x = 1; x <= max_numbers; x++) {
    rand = Math.floor(Math.random() * numbers.length);
    num = numbers[rand];
    nums.push(num);
    numbers.splice(rand, 1);
  }

  document.querySelector('.loti').innerHTML = nums.join(",")
}
lottoGame(5)

<div class="loti"></div>

Unique (non-repeating) random numbers in O(1)?

Initialize an array of 1001 integers with the values 0-1000 and set a variable, max, to the current max index of the array (starting with 1000). Pick a random number, r, between 0 and max, swap the number at the position r with the number at position max and return the number now at position max. Decrement max by 1 and continue. When max is 0, set max back to the size of the array - 1 and start again without the need to reinitialize the array.

Update:
Although I came up with this method on my own when I answered the question, after some research I realize this is a modified version of Fisher-Yates known as Durstenfeld-Fisher-Yates or Knuth-Fisher-Yates. Since the description may be a little difficult to follow, I have provided an example below (using 11 elements instead of 1001):

Array starts off with 11 elements initialized to array[n] = n, max starts off at 10:

+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|10|
+--+--+--+--+--+--+--+--+--+--+--+
                                ^
                               max    

At each iteration, a random number r is selected between 0 and max, array[r] and array[max] are swapped, the new array[max] is returned, and max is decremented:

max = 10, r = 3
           +--------------------+
           v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 7| 8| 9| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 9, r = 7
                       +-----+
                       v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 9| 8| 7: 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 8, r = 1
     +--------------------+
     v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 5| 6| 9| 1: 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 7, r = 5
                 +-----+
                 v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 9| 6| 5: 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

...

After 11 iterations, all numbers in the array have been selected, max == 0, and the array elements are shuffled:

+--+--+--+--+--+--+--+--+--+--+--+
| 4|10| 8| 6| 2| 0| 9| 5| 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

At this point, max can be reset to 10 and the process can continue.

Generate a non-repeating random number in JavaScript

You have 2 mistakes, oné is the array inside the function this cleared for each try, and then there is wrong logic ending up in an infinite loop.

const usedIndexes = [];    
function getUniqueRandomNumber(x) {
  const index = Math.floor(Math.random() * (x));
  if (usedIndexes.includes(index)) {
    return this.getUniqueRandomNumber(x);
  } else { 
    console.log(index);
    usedIndexes.push(index);
    return index;
  }
}

Also, I would think about using Set, in this situation instead of the array.

const usedIndexes = new Set();    
function getUniqueRandomNumber(max, min = 0) {
  const newNumber = Math.floor(Math.random() * (max - min) + min);
  if (usedIndexes.has(newNumber)) {
    return this.getUniqueRandomNumber(max, min);
  } else { 
    usedIndexes.add(newNumber);
    return newNumber;
  }
}

I have also edited variables names to better reflect their actual use and added a minimum for a random number.

JavaScript generate random numbers without repeating

use a variable to check if the number is the same.

something like this: (using LastNumber to store the lastNumber) if it allready is used we gonna try again)

var videoLinks = [
    ....
];

var lastNumber = 0;
var randomNumber = function () {
    var getRandomNumber = Math.floor(Math.random() * 5);  
    if(getRandomNumber != lastNumber){
        var random = videoLinks[getRandomNumber];
        document.getElementById("videoWrapper").innerHTML = random[0];
        lastNumber = getRandomNumber;
    }else{
        randomNumber();
    }
};

randomNumber(); // To call the function on load

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