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.

Solution1: 

非常strightforward，算和，看是不是single digit。

public class Solution {
    public int addDigits(int num) {
        while(num/10!=0)
        {
            num=sum(num);
        }
        return num;
    }

    public int sum(int num)
    {
        int res=0;
        while(num!=0)
        {
            res+=num%10;
            num/=10;
        }
        return res;
    }
    
}

Solution2: 

follow up: O(1)

ref:https://en.wikipedia.org/wiki/Digital_root

有规律的。。。除9余数为0，树根为9.除9余数不为零，为余数。0的树根为0.

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