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

 1     public int addDigits(int num) {
 2         while (num/10!=0)
 3         {
 4             int temp = 0;
 5             while (num/10!=0)
 6             {
 7                 temp+=num/10;
 8                 num %=10;
 9             }
10             temp += num;
11             num = temp;
12         }
13         return num;        
14     }

258. Add Digits

原文：http://www.cnblogs.com/wzj4858/p/7701392.html