258. 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?

题目含义：给定一个非负整数，每一位上的数字求和后得出新的整数，重复此过程，返回只有一位时候的结果

    public int addDigits(int num) {
        while ( num / 10 != 0 )
            num = num/10 + num%10;
        return num;
    }