Add Digits (数根 digital root)

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?

循环求解的方法就不介绍了。

一个数循环相加得到的个位数(digital root, 数根) = 这个数 mod 9

digital root的介绍详见https://en.wikipedia.org/wiki/Digital_root

so

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

简化版

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