【LeetCode】258. Add Digits 解题小结

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.

参考这一篇wiki百科

https://en.wikipedia.org/wiki/Digital_root

class Solution {
public:
    int addDigits(int num) {
        if (num >= 0 && num < 10) return num;
        
        return (num - 9 * ((num-1)/9));
    }
};

【LeetCode】258. Add Digits 解题小结

原文：http://www.cnblogs.com/Doctengineer/p/5860940.html