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.

//给定一个非负整数num，重复地将其每位数字相加，直到结果只有一位数为止。
class Solution{
public:
	int addDigits(int num){
		while (num > 9){
			int c = 0;
			while (num > 0){
				c = c + num % 10;
				num = num / 10;
			}
			num = c;
		}
		return num;
	}
};