LeetCode:Add Digits

fullstack_lth

于 2015-10-26 23:16:55 发布

阅读量414

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： leetcode c++

本文链接：https://blog.youkuaiyun.com/u014744118/article/details/49431805

161 篇文章

订阅专栏

本文介绍了一种高效的算法，用于计算任意非负整数的数字根，即反复加总该整数的所有位直到只剩一位数的过程。文章提供了一种不依赖循环或递归的O(1)运行时解决方案，并附带了简洁的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

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:

Show Hint

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

思路：

digit = num - 9 * ((n - 1) / 9)

代码：

class Solution {
public:
    int addDigits(int num) {
    int  digit = num - 9 *((num - 1) / 9);
    return digit;
    }
};