function numberSum(num) {
var str = num.toString();
var arrNum = str.split('').map(Number);//arrNum = [1, 2, 3];
//For-looping
var result = 0;
for (var i = 0; i < arrNum.length; i++) {
result = result + arrNum[i];
}
return result;
}
console.log(numberSum(22222)); // 2 + 2 + 2 + 2 + 2 = 10I did this with For-looping and then iterate it. The question is, how do i did the same but with Recursive Function?
You could use only the first element for addition and call the function with the rest of the array again.
In this case, a check is made for the length, this returns either 0 if the array has no items or the item count, then a shift is made which returns the first item of the array. Additionaly the function is called again with the reduced array.
function iter(array) { return array.length && array.shift() + iter(array); // ^^^^^^^^^^^^ exit condition, // if zero, return zero, // otherwise return the // ^^^^^^^^^^^^^^^^^^^^^^^^^^^ iteration part // return the first value and // call recursion again }
function numberSum(v) {
function iter(array) {
return array.length && array.shift() + iter(array);
}
return iter(v.toString().split('').map(Number));
}
console.log(numberSum(22222)); // 2 + 2 + 2 + 2 + 2 = 10For the input you have (22222) your function is a practical solution. If you want a function that takes a number and adds itself together a certain number of times you can simply do this...
function sums(a, b) {
return a * b;
}
sums(2, 5);
//=> 10
But if you really require an example of a recursive function to do this the following will achieve the same result...
var num = 2;
var iterate = 5;
function sums(n, count, total) {
if (count === 0) {
return total;
} else {
return sums(n, --count, total+n);
}
}
console.log(sums(num, iterate, 0));
//=> 10
Hope that helped. :)
(See this blog post on JavaScript recursion by integralist).
