258. Add Digits

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Total Accepted: 67030
Total Submissions: 139534
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?

Hint:

A naive implementation of the above process is trivial. Could you come up with other methods?

O(1)的话自然不能用循环了，答案就是余九之后的结果，但除了0之外9的倍数的话都还是9，所以改动一下减1再余九再加一即可。

C++:

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class Solution {
public:
    int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
};

有趣的是同样的代码Java比C++和C都跑得快

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