leetcode 258. Add Digits

最新推荐文章于 2025-08-11 15:06:51 发布

sonchcng

最新推荐文章于 2025-08-11 15:06:51 发布

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文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/sonchcng/article/details/51302372

本文介绍了一个简单的算法问题：给定一个非负整数，重复将其各位数字相加直至只剩一位数字，并返回该数字。提供了高效的解决方案，探讨了如何在O(1)运行时间内不使用循环或递归解决此问题。

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My Submissions QuestionEditorial Solution
Total Accepted: 91671 Total Submissions: 189101 Difficulty: Easy
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?

Show Hint

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