Add Digits

题目如下：
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?

这也就是一个数的Digital root。结论是

简单的一个思路是N=10ⁿ * a_n + 10^n-1 * a_n-1 + … + 10 * a₁ + a₀ =
(a_n + a_n-1 + … +a₁ + a₀) + (9ⁿ * a_n + 9^n-1 * a_n-1 + … + 9 * a₁ + a₀)
其中后者整除9则Nmod9和 (a_n + a_n-1 + … +a₁ + a₀) mod9 结果一样。直接根据结论写出代码不难，算法如下：

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