LeetCode 258. Add Digits （int转string）

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?

输入一个整数，计算其各位之和，如此迭代直到和为个位数。

思路：int转string—— int to_string(int) 函数

    int addDigits(int num) {
        int sum=num;
        while(sum>=10)
        {
            string str = to_string(sum);
            sum=0;
            for(int i=0;i<str.size();i++)
                sum+=int(str[i]-'0');
        }
        return sum;
    }

挑战：不用循环实现

    int addDigits(int num) {
        return 1 + (num - 1) % 9;
    }