leetcode_java_adddigits

题目如下：

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?

解析：

122  = 1+2+2;

122 = 122-(100-1+20-2)

40=40-4

50=50-5

...

可以看出是去除掉所有的9的倍数，故是对算法除9取模。同时考虑等于0的时候的除9与大于0的时候除9，代码如下：

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