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.

题意：将给定的整数各个位数相加，直到相加之和小于10。

思路：（1）常规的解法，求和之后进行判断，然后在进行求和直至满足条件；（2）参考的定义进行求解。

代码：

第一种思路：

//求整数的各位之和        public int get_sum(int sum)        {         int count = 0;         while(sum>0){             count = count + sum%10;             sum/=10;         }         return count;        }     public int addDigits(int num) {         int sum=0;          while(true)          {              sum= get_sum(num);              if(num<=9)                 {                      return sum;                }              else               {                num = sum;              }          }      }

第二种思路：

public int addDigits(int num) {            if (num == 0) return 0;            if (num%9==0) return 9;            if(num<9) {                return num;            }else{                return addDigits(num % 9);            }        }