258. Add Digits

258 - Add Digits
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Question
Editorial Solution
Total Accepted: 101723 Total Submissions: 207987 Difficulty: Easy

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
Hint:
A naive implementation of the above process is trivial. Could you come up with other methods?
What are all the possible results?
How do they occur, periodically or randomly?
You may find this Wikipedia article useful.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

public class Solution {
    public int addDigits(int num) {
        /*
        https://en.wikipedia.org/wiki/Digital_root
        dr(n)  = 1 + ((n - 1) mod 9)
        e.g. dr(371)  =   3* (100%9) + 7 * (10%9) + 1%9 = (3+7+1) % 9  = 2
        */
        int digitRoot =  1 + (( num - 1) %9);
        
        return digitRoot;
    }
    
    public int addDigits_recursive(int num) {
        int res = 0;
        while (num >= 10) {
            res += num % 10;
            num = num / 10;
        }
         res = res + num;
         return res >= 10 ?  addDigits(res) : res;
    }
}