题目描述

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?

解法一

思路

这题没啥意思，算是一个经典的数学问题，叫做Digital Root，其解就是f(n) = 1 + (n - 1) % 9。

Python

class Solution(object):
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        if num == 0:
            return 0
        return 1 + (num - 1) % 9

Java

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

C++

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