Leetcode ☞ 258. Add Digits

网址：https://leetcode.com/problems/add-digits/

258. Add Digits

My Submissions

Question

Total Accepted: 72085 Total Submissions: 149892 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?

我的AC：

int addDigits(int num) {    if(num)        return num % 9 ? num%9 : 9;    else        return 0;}

分析：

不让用循环，说明有一定规律，观察即得。

测试数据：0 1 8 9 10 18 27 32

即输出可分为3类（两种分法）：0；9的倍数；非9的倍数。或者小于10，输出num；大于10的9的倍数；大于10的非9倍数。