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:

3 methods for python with explainsIteration method
      class Solution(object):      def addDigits(self, num):        """        :type num: int        :rtype: int        """        while(num >= 10):            temp = 0            while(num > 0):                temp += num % 10                num /= 10            num = temp        return num
Digital Rootthis method depends on the truth:N=(a[0] * 1 + a[1] * 10 + ...a[n] * 10 ^n),and a[0]...a[n] are all between [0,9]we set M = a[0] + a[1] + ..a[n]and another truth is that:1 % 9 = 110 % 9 = 1100 % 9 = 1so N % 9 = a[0] + a[1] + ..a[n]means N % 9 = Mso N = M (% 9)as 9 % 9 = 0,so we can make (n - 1) % 9 + 1 to help us solve the problem when n is 9.as N is 9, ( 9 - 1) % 9 + 1 = 9
class Solution(object):def addDigits(self, num):    """    :type num: int    :rtype: int    """    if num == 0 : return 0    else:return (num - 1) % 9 + 1