Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.
Example 1:
Input: num = 38 Output: 2 Explanation: The process is 38 --> 3 + 8 --> 11 11 --> 1 + 1 --> 2 Since 2 has only one digit, return it.
Example 2:
Input: num = 0 Output: 0
Constraints:
0 <= num <= 231 - 1
Follow up: Could you do it without any loop/recursion in O(1) runtime?
class Solution:
def addDigits(self, num: int) -> int:
return 0 if num == 0 else (num - 1) % 9 + 1class Solution {
public int addDigits(int num) {
return (num - 1) % 9 + 1;
}
}class Solution {
public:
int addDigits(int num) {
return (num - 1) % 9 + 1;
}
};