Description：
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38
Output: 2

Explanation: 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?

题意：给定一个非负的整数，将这个数的各个位相加知道最终的结果是一位时停止，并返回这个数；并且不使用循环和递归；

解法：如果这个数是0的话，那么肯定就是0了；我们发现如果这个数是9的倍数的话，那么结果就是9；否则就是对9取余；

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