题目描述

题目链接：258. Add Digits

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

例子

Input: 38
Output: 2
Explanation: 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?

解题思路

简单地通过循环相加各个位直到最终 num < 9 即可，代码如下：

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class Solution {
public:
    int addDigits(int num) {
        int result = 0;
        while (num > 9) {
            int next_num = 0;
            while (num > 0) {
                next_num += (num % 10);
                num /= 10;
            }
            num = next_num;
        }
        return num;
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(1)

GitHub 代码同步地址： 258.AddDigits.cpp

其他题目：
GitHub: Leetcode-C++-Solution
博客： Leetcode-Solutions