# Add Digits

<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?

## Thoughts

都写下来会发现有规律. 以1\~9循环. 由于9%9 == 0, 所以让(num - 1) % 9 + 1.

## Code

class Solution { public int addDigits(int num) { return (num - 1) % 9 + 1; } }

## Analysis

时空复杂度O(1).