20160307 Add Digits

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：
Significance and formula of the digital root（digital root 公式）：

代码：

public class Solution {    public int addDigits(int num) {        int temp = (num - 1) / 9 * 9;        return (num - temp);    }}