258.Add Digits
===

###### tags: `Easy`,`Math`

[258. Add Digits](https://leetcode.com/problems/add-digits/)

### 題目描述

Given an integer `num`, repeatedly add all its digits until the result has only one digit, and return it.

### 範例

**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` <= 2^31^ - 1

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

### 解答
#### Javascript

function addDigits(num) {
  if (num < 10) return num;
  return addDigits((num % 10) + addDigits(Math.floor(num / 10)));
}

> [name=Marsgoat][time=Apr 26, 2023]

function addDigits(num) {
  return ((num - 1) % 9) + 1;
}

> 吉神：這題O(1)不是基本嗎？9的倍數檢查方式不是國小教的嗎？？
> 我道歉嗚嗚嗚，我就爛。
> [name=Marsgoat][time=Apr 26, 2023]

#### C++

class Solution {
public:
    int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
};

> [name=Yen-Chi Chen][time=Wed, Apr 26, 2023]

#### Python

class Solution:
    def addDigits(self, num: int) -> int:
        return (num - 1) % 9 + 1 if num else 0

> [name=Yen-Chi Chen][time=Wed, Apr 26, 2023]

#### C#

public int AddDigits(int num) {
    if (num == 0) return 0;
    int r = num % 9;
    return r == 0 ? 9 : r;
}

> [name=Jim][time=Wed, Apr 26, 2023]
\begin{split}
&x_n10^n+x_{n-1}10^{n-1} +...+x_110+x_0\\&=x_n(9+1)^n+x_{n-1}(9+1)^{n-1} +...+x_1(9+1)+x_0\\&=9(...) +(x_n+x_{n-1}+...+x_1 +x_0)
\end{split}
忘了怎麼排版...

#### Java

class Solution {
    public int addDigits(int num) {
        return num != 0 ? (num - 1) % 9 + 1 : 0;
    }
}

> [name=Ron Chen][time=Mon, Jun 12, 2023]
### Reference

[排版教學](https://hackmd.io/@sysprog/B1RwlM85Z?type=eidt#LaTeX-%E6%8E%92%E7%89%88)

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