LeetCode 258. Add Digits

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, 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?

class Solution {// 水题
public:
    int addDigits(int num) {  
        while(num>=10){
            int temp=0;
            while(num!=0){
                temp+=num%10;
                num/=10;
            }
            num=temp;
        }
        return num;
    }
};
原文地址:https://www.cnblogs.com/A-Little-Nut/p/10058649.html