使用递归对数组中的元素求和，即使它们是字符串

Question

Have to create a function that return the sum of the element in the array but if the array is

["a","b","c"] // output : abc    

So far I have

 function calculateSumRecursion(array) {
          //your code
          if (array.length === 0 ) {
            return 0
          }
          
      return array[0] + calculateSumRecursion(array.slice(1))
    }

I found out how to calculate the sum of all numbers using recursion but when it's an array of string like

array = ["a","b","c"] 

it returns me

// abc0

because of the if statement.. is there any way to say

if (array.length === 0) return nothing instead of a 0 (that work only when it's an array of number?)

Answer 1

You just need to return the only value in the array when the length is 1, rather than waiting until you get a length of 0. That way you are always summing compatible types (numbers or strings). Note that you still need a test for a 0 array length in case the function gets called with an empty array. In this case you need to choose what to return; as requested, it is 0.

 function calculateSumRecursion(array) { if (array.length === 0) { return 0; } if (array.length === 1) { return array[0]; } return array[0] + calculateSumRecursion(array.slice(1)) } console.log(calculateSumRecursion([1, 2, 3, 4, 5])); console.log(calculateSumRecursion(['a', 'b', 'c'])); console.log(calculateSumRecursion([]));

Answer 2

let arr = [1,2,3,4,5] // output : abc
let sum = calculateSumRecursion(arr);

function calculateSumRecursion (arr) {
    return arr.length ? arr.pop() + calculateSumRecursion(arr) : 0;
}

Slice version

let arr = [1,2,3,4,5] // output : abc
let sum = calculateSumRecursion(arr);

function calculateSumRecursion (arr) {
    return arr.length ? arr[0] + calculateSumRecursion(arr.slice(1)) : 0;
}

Answer 3

Return empty string on recursion base case. Just replace your return 0 to return '' .

 const array = ['a', 'b', 'c']; function calculateSumRecursion(array) { if (array.length === 0) { return ''; } return array[0] + calculateSumRecursion(array.slice(1)); } console.log(calculateSumRecursion(array));

If you are want to work with number also then check array length for zero as well as one.

 const array = ['a', 'b', 'c', 'e']; const array2 = []; const array3 = [1, 2, 3]; function calculateSumRecursion(array) { const rec = array.length === 1? array[0]: array.length >= 1 && array[0] + calculateSumRecursion(array.slice(1)); return array.length === 0? 0: rec; } console.log(calculateSumRecursion(array)); console.log(calculateSumRecursion(array2)); console.log(calculateSumRecursion(array3));

Answer 4

Change return 0 to return "" which will add an empty string to the sum.

Answer 5

You have returned 0 when the array is empty.

Now, you are doing string operations so it is needed to return empty value (not zero) so it will be affordable to return "" .

function calculateSumRecursion(array) {
  return array.length === 0 ? "" : array[0] + calculateSumRecursion(array.slice(1));
}

Answer 6

There's a way easier way to do this:

    function calculateSumRecursion(array) {
        var out = array[0];
        for (let i = 1; i < array.length; i++) {
            out = out + array[i];
        }
        return out;
    }