Challenge from 10_04_2018

First solution

function challenge1 (a, b, c){
    let base1 = Number.parseInt(a, b);
    let base2 = Number.parseInt(base1, c);
    return base2.toString();
}

console.log(challenge1 (10, 2, 10))

Second solution

function base2base (n, base1, base2){
    var ourNumber = 0;
    for (var number_position= n.length - 1, i = 0; number_position >= 0; -- number_position, i++){
        ourNumber += n[number_position]*Math.power(base1, i)
    }
    return ourNumber.toString(base2);
}


console.log(base2base (10, 2, 10))



*/

Podejscie proste wbudowane:
znajdz w dokumentacji opis funkcji parseInt i wykorzystaj odpowiednio

Podejscie iteracyjne
function base2base(n, base1, base2) {
	var liczba = 0
	for number_position = n.length - 1, iterator = 0 ; number_position >= 0 ; --number_position, ++iterator {
		liczba += n[number_position] * (base1 ** iterator)
	} 
	return liczba.toString(base2)
}

n = 10011 = 1 * 2**4 + 0 * 2 ** 3 + 0 * 2 ** 2 + 1 * 2 ** 1 + 1 * 2 ** 0 = 19
b1 = 2
b2 = 10

1 loop:
number_position = 5 - 1 = 4
iterator = 0
n[number_position] = 1
base1 ** iterator = 2 ** 0 = 1
liczba = 0 + 1

2 loop:
number_position = 3
iterator = 1
n[number_position] = 1
base1 ** iterator = 2 ** 1 = 2
liczba = 1 + 2 = 3

3 loop:
number_position = 2
iterator = 2
n[number_position] = 0
base1 ** iterator = 2 ** 2 = 4
liczba = 3 + 0

4 loop:
number_position = 1
iterator = 3
n[number_position] = 0
base1 ** iterator = 2 ** 3 = 8
liczba = 3 + 0

5 loop:
number_position = 0
iterator = 4
n[number_position] = 1
base1 ** iterator = 2 ** 4 = 16
liczba = 3 + 16 = 19


*/