258. 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?

一刷
题解：Digit Root解法
如果进制为b(这道题中为10），输入为n, digit root of n is:

• dr(n) = 0 if n==0
• dr(n) = (b-1) if n!=0 and n%(b-1) == 0
• dr(n) = n % (b-1) if n % (b-1)!=0
  或者：
• dr(n) = 1 + (n-1) % (d-1)
  因为：

dr(abc) = (a*10^2 + b*10 + c)%9 = (a+b+c)%9

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