[leetcode] 258. Add Digits 解题报告

题目链接：https://leetcode.com/problems/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?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

思路：这是一道数学可以解的问题，只需要一个公式，value = 1 + (n-1)%9; 也就是说这题只需要一行代码，可是我并不高兴，因为公式的推导好长，我看不懂，也不想看。就在写一个不断除余分解的吧！

代码如下：

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