(leetcode) Add Digits---integers everybody add

Given a non-negative integer num , repeatedly add all its digits until the result have only one digit.

For example:

Givennum = 38, the process is like:3 + 8 = 11,1 + 1 = 2. Since2have only one digit, return it.

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

Problem Solving Analysis:

Solution One:

The remainder is added to Num, and the remainder is added until NUM is less than two digits.

#-*-Coding:utf-8-*-__author__ = ' Jiuzhang ' class solution (object):    def adddigits (self, num): While        num >= 10 :            num = (NUM/10) + num%        return num

Solution Two:

The other method is relatively simple and can be illustrated by examples. Assuming that the number entered is a 5 digit num,

Num's members are A, B, C, D, E, respectively. Has the following relationship: num = A * 10000 + b * + + c * + + D * + E

That is: num = (A + B + C + D + E) + (A * 9999 + b * 999 + c * + D * 9)

Because A * 9999 + b * 999 + c * + D * 9 must be divisible by 9, so the num modulo results in addition to 9

The result is the same as a + B + C + D + E modulo except 9. Perform the same operations on the number A + B + C + D + E repeatedly,

The final result is a 1-9 number plus a bunch of numbers, the leftmost number is 1-9, the right number is always

Can be divisible by 9.

Attention:

To handle special cases, when the number is 0, please return directly to 0

#-*-Coding:utf-8-*-__author__ = ' Jiuzhang ' class solution (object):    def adddigits (self, num):        if num > 0:
   
    return 1 + (num-1)% 9        else:            return 0
   

(leetcode) Add Digits---integers everybody add