leetcode258

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1、Add Digits
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?

class Solution {public:    int addDigits(int num) {        if(num!=0&&num%9==0)            return 9;        return (num%9);    }};

任意给一个正整数，每一位上的数相加获得一个新的整数，然后接着对这个整数，每一位上的数相加，直到成为个位数，则这个个位数一定是最初给的正整数整除9 的余数。。。

number                   result   0                     0  // 0 % 9 = 0, 0 == 0   1                     1 // 1 % 9 = 1   2                     2   3                     3   4                     4   5                     5   6                     6   7                     7   8                     8   9                     9 // 9 % 9 = 0, 9 != 0   10                    1 // 10 % 9 = 1   11                    2   12                    3   13                    4   14                    5   15                    6   16                    7   17                    8   18                    9   19                    1 //19 % 9 = 1

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