第十周作业： Add Digits

题面：

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?

题解：将每个数的各个位上的数字加起来，若大于9，则继续加，个位数则输出。则自然可想到，模除和累加，一个while即可。做完后觉得很笨。思考后则可得任何大于9得数按此法即是求对9的模。则可更简便。

class Solution {public:    int addDigits(int num) {            while(num>9){                 int sum=0;                 while(num){                  sum+=num%10;                  num/=10;                 }                  num=sum;                 }                 return num;    }};

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