[leetcode] 258. Add Digits 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/50412421

本文探讨了LeetCode上的一道经典题目——加数问题，给出了一种仅需一行代码的解决方案，并解释了背后的数学原理。同时提供了一个通过将数字转换为字符串进行处理的示例代码。

题目链接：https://leetcode.com/problems/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?

Hint:

A naive implementation of the above process is trivial. Could you come up with other methods?
What are all the possible results?
How do they occur, periodically or randomly?
You may find this Wikipedia article useful.

思路：这是一道数学可以解的问题，只需要一个公式，value = 1 + (n-1)%9; 也就是说这题只需要一行代码，其原理是在数学中任何可以被9整除的数, 其数字相加同样可以被9整除. 而要求的结果范围同样是[0, 9], 因此问题就转化为除余的问题.

就再写一个转换成字符串处理的吧.

代码如下：

class Solution {
public:
    int addDigits(int num) {
        string tem = to_string(num);
        while(tem.size() > 1)
        {
            int sum = 0;
            for(auto val: tem) sum += val-'0';
            tem = to_string(sum);
        }
        return stoi(tem);
    }
};