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.

Follow up: Could you do it without any loop/recursion in O(1) runtime?

class Solution(object):
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        while num / 10 >= 1:
            new = 0
            for c in str(num):
                new += int(c)
            num = new
        return num

O(1)

抄来的:

this 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 = 1
10 % 9 = 1
100 % 9 = 1
so N % 9 = a[0] + a[1] + ..a[n]
means N % 9 = M
so 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