Add Digits

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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?

Analyse: 

1. trivival solution:

 1 class Solution {
 2 public:
 3     int addDigits(int num) {
 4         while(num >= 10) num = add(num);
 5         return num;
 6     }
 7     int add(int num){
 8         int result = 0;
 9         while(num){
10             result += num % 10;
11             num /= 10;
12         }
13         return result;
14     }
15 };

we can find the result after implementation:

0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3
4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7
8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 请按任意键继续. . .

so the result is (num - 1) % 9 + 1;

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

Add Digits

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原文地址：http://www.cnblogs.com/amazingzoe/p/4762241.html