Leetcode 258

258. Add Digits

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Question

Total Accepted: 70241  Total Submissions: 146130  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?

刚开始也并不知知道如何O（1）的时间解决 ， 后来查找才知道原来有个数根公式

    digits = 1 + ( n - 1 )%9 ;

所以问题就简单了 ，一行 代码就搞定了 。

int addDigits(int num) {
       return (num-1)%9+1 ;
}