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?

给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。

• 给定一个非负整数，反复加各个位上的数，直到结果为一位数为止，最后输出这一位数。
• 简单题。按照题意循环累加即可。

package leetcode funcaddDigits(numint) int { fornum>9 { cur:=0fornum!=0 { cur+=num%10num/=10 } num=cur } returnnum }