# 258. Add Digits
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
Example:
Input: 38
Output: 2
Explanation: 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
Approach 1: Loop.
Approach 2: O(1).
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
class Solution {
public:
int addDigits(int num) {
int sum = 0;
while (true)
{
sum += num % 10;
num = num / 10;
if (num <= 0)
{
if (sum < 10)
break;
num = sum;
sum = 0;
}
}
return sum;
}
};
Approach 2:
class Solution {
public:
int addDigits(int num) {
if (num == 0) return 0;
if (num % 9 == 0) return 9;
return num % 9;
}
};
class Solution {
public:
int addDigits(int num) {
return 1 + ((num - 1) % 9);
}
};