leetcode-258-Add-Digits

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描述

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.

分析

找数学规律的一道题，比方说38，是3+8，而前面的3是30%9得到的，而11，1+1，前面的1是10%9得到的，

解决方案1(C++)

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class Solution {
public:
    int addDigits(int num) {
        if(num%9 == 0 && num != 0){
            return 9;
        }else {
            return num%9;
        }
    }
};

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