258. Add Digits

题目

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?

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思路

九余数定理
input: 0 1 2 3 4 …
output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ….

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代码

class Solution {public:    int addDigits(int num) {        //九余数定理        //input: 0 1 2 3 4 ...        //output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....        if(num == 0)        {            return 0;        }        return 1 + (num-1)%9;    }};