LeetCode(258):Add Digits

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Digital root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);
            }
        }