258. Add Digits

题目描述：

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?

思路一：

数字根问题：https://en.wikipedia.org/wiki/Digital_root#Congruence_formula

class Solution {
    public int addDigits(int num) {
        return 1 + (num - 1) % 9;
    }
}

思路二：

class Solution {
    public int addDigits(int num) {
        return num % 9 == 0 ? (num == 0 ? 0 : 9) : (num % 9);
    }
}