LeetCode--Add Digits

题目:

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?

方法一:

    int addDigits(int num) {
            int sum = 0;
            while(num > 0)
            {
               sum += num%10;
               num /= 10;
            }
            if(sum < 10)
            {
                return sum;
            }
            else
            {
                addDigits(sum);
            }
    }

while循环中将一个数的每一位加到sum中,每次都是将个位加上去,再除以10.

再判断sum是否小于10,大于10则递归.

方法二:

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

如果能看出 最后的结果就是对9的余数,则只用O(1)的时间复杂度