Leetcode--add Digits

Description:

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?

Test instructions is very simple, the easiest way to think of is to loop the iteration, if it is <10 return. The following is a recursive approach.

public class Solution {public    int adddigits (int num) {        if (num <) {            return num;        }        1023        int t = 0;        while (num >= 1) {            t + = num%;            num = NUM/10;        }                return Adddigits (t);    }}

　　So what can I do to reduce the complexity to O (1) without recursion and iteration, which reminds me of the formula. To find patterns through data.

Top 30 Data tests:

public static void Main (string[] args) {for (int i=0; i<30; i++) {System.out.println (i + "+ adddigits (i))}}

　　

0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 1
11 2
12 3
13 4
14 5
15 6
16 7
17 8
18 9
19 1
20 2
21 3
22 4
23 5
24 6
25 7
26 8
27 9
28 1
29 2

Find out the rules.

In fact, this is: (num-1)% 9 + 1

public class Solution {public    int adddigits (int num) {       return (num-1)% 9 + 1;    }}

Leetcode--add Digits