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?

分析：求各位數之和，並要求加到只有一位數為止。

思路：因為是加法，很容易找規律，直接挨個列出結果。

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

基本可以確定規律 (num-1)%9+1

JAVA CODE

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

Add Digits