[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：

需要时间复杂度是O(1)，所以需要分析规律

经过计算1~30，得出结论：结果在1~9之间，提交代码为：

public static int addDigits(int num) {
int intRsult  = num%9;
if(intRsult == 0){
intRsult =9;
}
return intRsult;
}

但是结果报错，发现遗忘了输入为0的情况，修改代码为：

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