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?

思路：

1、常规解法：循环或者递归

2、数字根问题，参考维基百科的证明和结论：https://en.wikipedia.org/wiki/Digital_root#Congruence_formula.

For base b (decimal case b = 10), the digit root of an integer is:

• dr(n) = 0 if n == 0
• dr(n) = (b-1) if n != 0 and n % (b-1) == 0
• dr(n) = n mod (b-1) if n % (b-1) != 0

or

• dr(n) = 1 + (n - 1) % 9

代码1：

class Solution {
public:
    int addDigits(int num) 
    {
        if(num <= 0)
            return 0;
        while(num >= 10)
        {
            int tmp = 0;
            while(num >= 10)
            {
                tmp += num % 10;
                num = num / 10;
            }
            tmp += num;
            num = tmp;
        }
        return num;
    }
};

代码2：

class Solution {
public:
    int addDigits(int num) 
    {
        if(num <= 0)
            return 0;
        if(num % 9 != 0)
            return num % 9;
        else 
            return 9;
    }
};

代码3：

class Solution {
public:
    int addDigits(int num) 
    {
        if(num <= 0)
            return 0;
        return 1 + (num - 1 ) % 9;
    }
};

LeetCode OJ 之 Add Digits （数字相加）

原文：http://blog.csdn.net/u012243115/article/details/48133139