Codewars JS习题：Sum of Digits / Digital Root

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题目

In this kata, you must create a digital root function.

A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has two digits, continue reducing in this way until a single-digit number is produced. This is only applicable to the natural numbers.

Here’s how it works (Ruby example given):

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digital_root(16)
=> 1 + 6
=> 7

digital_root(942)
=> 9 + 4 + 2
=> 15 ...
=> 1 + 5
=> 6

digital_root(132189)
=> 1 + 3 + 2 + 1 + 8 + 9
=> 24 ...
=> 2 + 4
=> 6

digital_root(493193)
=> 4 + 9 + 3 + 1 + 9 + 3
=> 29 ...
=> 2 + 9
=> 11 ...
=> 1 + 1
=> 2

我的解答：

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function digital_root(n) {
  function sum(n) {
    let arr = [];
    let result = 0;
    n = n + '';
    for(let i of n) {
      result += parseInt(i);
    }
    return result;
  }
  let res = sum(n);
  if(res > 9) {
    return sum(res);
  } else {
    return res;
  }
}

思路:
因为结果需要一直重复轮加，直到最终结果小于10，所以会考虑到递归。在函数内部创建一个基础的函数，先算出n所有位数相加的结果，如果大于10，再对结果使用基础函数。
基础函数的思路是，先将参数数字n转换为字符串，使其具有可遍历性，然后轮加，求得最终结果，返回给digital_root函数。