leetcode:258 Add Digits－每日编程第二题

Add Digits

Total Accepted: 49005 Total Submissions: 102997 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:

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

思路：

假设给了一个4位数num,其值为ABCD，那么num=1000*A+100*B+10*C+D=A+B+C+D+999*A+99*B+9*C=A+B+C+D+9*(111*A+11*B+C)

那么num%9就等于A+B+C+D,而num%9<10,这么看num%9似乎就是我们要的结果。

但到这还没有完，当num=9时，由于9%9等于0，并不是我们想要的结果。因此，我们改进为(num-1)%9+1。

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