问题描述
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?
Hint:
A naive implementation of the above process is trivial. Could you come up with other methods?
代码
class Solution {
public:
int addDigits(int num) {
return (num - 1) % 9 + 1;
}
};