Add Digits

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题目：

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?

解析：

就是把一个整数的每个数字分离相加

 1 class Solution {
 2 public:
 3     int addDigits(int num) {
 4         if(num < 10)
 5             return num;
 6         else
 7         {
 8             int n = 0;
 9             while(num != 0)
10             {
11                 n += num%10;
12                 num = num/10;
13             }
14             addDigits(n);
15         }
16     }
17 };

Add Digits

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原文地址：http://www.cnblogs.com/raichen/p/4926332.html