Leetcode 258. ADD Digits

Problem:

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

For example:

Given num = 38 , the process is like: 3 + 8 = 11 , 1 + 1 = 2 . Since have only one 2 digit, return it.

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

The original idea is to add and judge the repeated loops until a satisfied solution is found. By referring to other people's answers, we find that the problem is to ask for "number root".

This method is judged by the cyclic addition until a viable solution is found:

classSolution { Public:    intAdddigits (intnum) {        intAdd =0;  while(Num >=Ten)        {             while(Num >0) {Add+ = num%Ten; Num/=Ten; } num=add; }        returnnum; }};

However, due to the complexity of time exceeds the requirements of the topic, it is considered to the "number root" solution.

The enumeration indicates that:
10 1+0 = 1
11 = 2
12 1+2 = 3
13 1+3 = 4
14 1+4 = 5
15 1+5 = 6
16 1+6 = 7
17 1+7 = 8
18 1+8 = 9
19 1+9 = 1
20 2+0 = 2

So we can draw a regular, every 9 cycles. To handle a multiple of 9, we write as (n-1)% 9 + 1. This can be done by:

class Solution {public:    int adddigits (int  num)     {         return 1 9 1 ;    }};

Leetcode 258. ADD Digits