LeetCode:Add Digits



问题描述：

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:

Show Hint

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

思路：

digit = num - 9 * ((n - 1) / 9)

参考自维基百科：https://en.wikipedia.org/wiki/Digital_root

代码：

class Solution {public:    int addDigits(int num) {    int  digit = num - 9 *((num - 1) / 9);    return digit;    }};