add-digits

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

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

&nbsp;
Example 1:

<pre>
Input: num = 38
Output: 2
Explanation: The process is
38 --&gt; 3 + 8 --&gt; 11
11 --&gt; 1 + 1 --&gt; 2 
Since 2 has only one digit, return it.
</pre>

Example 2:

<pre>
Input: num = 0
Output: 0
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>0 &lt;= num &lt;= 231 - 1</code></li>
</ul>

&nbsp;
Follow up: Could you do it without any loop/recursion in <code>O(1)</code> runtime?

addDigits

Add Digits

Happy Number

happy-number

快乐数

Sum of Digits in the Minimum Number

sum-of-digits-in-the-minimum-number

最小元素各数位之和

Sum of Digits of String After Convert

sum-of-digits-of-string-after-convert

字符串转化后的各位数字之和

Minimum Sum of Four Digit Number After Splitting Digits

minimum-sum-of-four-digit-number-after-splitting-digits

拆分数位后四位数字的最小和

Calculate Digit Sum of a String

calculate-digit-sum-of-a-string

计算字符串的数字和

Difference Between Element Sum and Digit Sum of an Array

difference-between-element-sum-and-digit-sum-of-an-array

数组元素和与数字和的绝对差

Alternating Digit Sum

alternating-digit-sum

交替数字和

Math

Number Theory

Simulation

给定一个非负整数 <code>num</code>，反复将各个位上的数字相加，直到结果为一位数。返回这个结果。

&nbsp;

示例 1:

<pre>
输入: num = <code>38</code>
输出: 2 
解释: 各位相加的过程为：
38 --&gt; 3 + 8 --&gt; 11
11 --&gt; 1 + 1 --&gt; 2
由于&nbsp;<code>2</code> 是一位数，所以返回 2。
</pre>

示例 2:

<pre>
输入: num = 0
输出: 0</pre>

&nbsp;

提示：

<ul>
	<li><code>0 &lt;= num &lt;= 231&nbsp;- 1</code></li>
</ul>

&nbsp;

进阶：你可以不使用循环或者递归，在 <code>O(1)</code> 时间复杂度内解决这个问题吗？