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## Problem – Add Digits LeetCode Solution

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?

## Add Digits LeetCode Solution in Java

``````class Solution {
if(num == 0) return 0;
else if(num % 9 == 0) return 9;
else return num % 9;
}
}
``````

## Add Digits LeetCode Solution in C++

``````class Solution {
public:
if(num == 0) return 0;
else if(num % 9 == 0) return 9;
else return num % 9;
}
};
``````

## Add Digits LeetCode Solution in Python

``````class Solution:
def addDigits(self, num: int) -> int:
if num == 0 : return 0
if num % 9 == 0 : return 9
else : return (num % 9)
``````
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