[leetcode] 258. Add Digits

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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?

递归，num/10 与 num％10

一刷：

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

[leetcode] 258. Add Digits

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原文地址：http://www.cnblogs.com/pulusite/p/5840614.html