258. Add Digits

看眼缘随机拿了一道不难的题https://leetcode.com/problems/add-digits/description/ 

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?

 提交测试通过：

int addDigits(int num) {
    if(num < 10) return num;
        int add = 0;
        while(num > 0){
            add += num % 10;
            num = num/10;
        }
        return addDigits(add);
}

用了递归，用循环做也行

258. Add Digits

原文：http://www.cnblogs.com/hozhangel/p/7745440.html