javascript odd and even separation in an array of numbers

Question

i wants to separate an array with two groups (odd and even) sequentially. but when i try this:

Answer 1

Because you move every found even value to the end of the array:

0: 1 2 3 4 5 6 7 8 9
     ^-------------v
1: 1 3 4 5 6 7 8 9 2   3 is not checked, because of the incremented index after splicing
       ^-----------v
2: 1 3 5 6 7 8 9 2 4   5 is not checked
         ^---------v
3: 1 3 5 7 8 9 2 4 6   7 is not checked
           ^-------v
4: 1 3 5 7 9 2 4 6 8   9 is not checked
             ^-----v
5: 1 3 5 7 9 4 6 8 2
               ^---v
6: 1 3 5 7 9 4 8 2 6
                 ^-v
7: 1 3 5 7 9 4 8 6 2
                   |
8: 1 3 5 7 9 4 8 6 2

But, you do not check the value after the found even value, like the value 3 in line zero, because it is never checked and stays at this place, as well as other actually uneven values. You could try the whole again with all even at the beginning of the array, like

var array = [2, 4, 6, 8, 1, 3, 5, 7, 9];
    
for (var i = 0; i < array.length; i++) {
    if (array[i] % 2 === 0) {
        array.push(array.splice(i, 1)[0]);
        console.log(i + ': ' + array.join(' '));
    }
}

console.log(array);

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Solution

You could use a length for checking and reduce the length for every found even value. In this case, the index stays at the value, because at the index is now the next element for checking.

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9],
    i = 0,
    l = array.length;

while (i < l) {
    if (array[i] % 2 === 0) {
        array.push(array.splice(i, 1)[0]);
        l--;
        continue;
    }
    i++;
}

console.log(array);

Or just sort.

By moving odd values to top and then sort by value.

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9];

array.sort((a, b) => b % 2 - a % 2 || a - b);
console.log(array);