Digit Sum in C++

Given a non-negative integer n, return the sum of its digits.

The digits are the figures from 0 to 9 that n is written with: 472 has the digits 4, 7 and 2. If n is 0, the sum is 0.

The last digit of a number is the remainder of its division by 10 (472 % 10 is 2), and dividing by 10 without the remainder removes that digit (472 becomes 47).

Example 1:

Input: n = 472
Output: 13
Explanation: Taking the digits from the right: 2 + 7 + 4 = 13.

Example 2:

Input: n = 1005
Output: 6
Explanation: 5 + 0 + 0 + 1 = 6. The zeros add nothing.

Example 3:

Input: n = 7
Output: 7
Explanation: A number with one digit is its own digit sum.

Constraints:

• 0 <= n <= 109