leetcode 258. Add Digits

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本文介绍了一个数字累加算法，该算法能将一个非负整数的所有位数不断相加直至结果仅剩一位数。文中提供了两种实现方式：一种是通过循环进行逐位累加；另一种则是利用数学规律直接计算结果，后者能在常数时间内完成运算。

258. Add Digits

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?

方法二是网上找的。找规律的方法确实吓人。

class Solution {
public:
    int addDigits(int num) 
    {
        //way-1
        /*
        while (num / 10 != 0)
        {
            int temp = 0;
            while (num / 10)
            {
                temp += num % 10;
                num /= 10;
            }
            temp += num;
            num = temp;
        }
        return num;
        */
        
        //way-2 找数据规律
        if (num == 0) 
            return 0;  
        return (num-1) % 9 + 1; 
    }
};