渣渣写LEETCODE——258. Add Digits

Problem:

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?

Solution:

用模10取出个位，用除10丢掉个位。

O(1)的解法想出来再补充。

Code:

/*__xz__*/
class Solution {
public:
int addDigits(int num) {
int result = 0;
if (num < 10) return num;
else {
while (num != 0) {
result += num%10;
num /= 10;
}
}

return result;
}
};