LeetCode 258. Add Digits 题解（C++）

LeetCode 258. Add Digits 题解（C++）

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题目描述

• Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

示例

• Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

限制条件

• Could you do it without any loop/recursion in O(1) runtime?

思路

• 这道题使用数根的公式，就可以达到O(1)的时间复杂度。
• 数根的公式如下:
  
  简化后的结果为
  有关数根的详细资料可查询https://en.wikipedia.org/wiki/Digital_root

代码

我的代码

class Solution
{
public:
    int addDigits(int num) 
    {
        while (num / 10 != 0)
        {
            num = (num / 10) + (num % 10);
        }
        return num;
    }
};

使用数根公式的代码

class Solution
{
public:
    int addDigits(int num) 
    {
        return 1 + (num - 1) % 9;
    }
};