LeetCode *** 258. Add Digits

题目：

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：

如果时间复杂度要求是O（1），那么可以进行如下分析：

数字	结果
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

那么代码如下：

int addDigits(int num) {
if(num==0)return 0;
if(num%9)return num%9;
else return 9;
}

还有一个代码更加简单，如下:

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