258. Add Digits

Problem

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?
Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.
   本质是数学问题，数根： 一个数的数根(digital root)就是它对9的余数
   数根就是不断地求这个数的各位数之和，直到求到个位数为止。所以数根一定和该数模9同余 ，但是数根又是大于零小于10的，所以数根模9的余数就是它本身，也就是说该数模9之后余数就是数根。 特殊情况是数根是9的时候，这时候余数是0
   12345 = 1 * 9999 + 2 * 999 + 3 * 99 + 4 * 9 + 5 + (1+ 2+ 3 + 4)
   solution

class Solution(object):
def addDigits(self, num):
return num if num == 0 else num % 9 or 9
使用递归，时间复杂度高于题目要求
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
j = 0
if num < 10:
return num
for i in str(num):
j += int(i)
return self.addDigits(j)