## 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:**

**Follow up:** Could you do it without any loop/recursion in `O(1)`

runtime?

## Add Digits LeetCode Solution in Java

```
class Solution {
public int addDigits(int num) {
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:
int addDigits(int num) {
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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