LeetCode Add Digits

原题链接在这里：https://leetcode.com/problems/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?

题解：

最基本的想法就是每一位相加知道成为个位数，但本题有follow-up, require O(1) time.

最后的结果只能是0-9. 列数看结果，发现num%9对于大多数正数得到了最后结果。

但corner case 9和9的倍数，本应该返回9, 却返回了0, 所以单独讨论。

Method 3 是返回(num-1)%9+1,  避免了9与9的倍数corner case, 但需要注意新的corner case num == 0. num - 1 就成了负数，此时num==0需要单独讨论。

Time Complexity: Method 3, O(1). Space: O(1).

AC Java:

public class Solution {
    public int addDigits(int num) {
        /*
        //Method 1
        while(num/10 != 0){
            int sum = 0;
            while(num>0){
                sum += num%10;
                num = num/10;
            }
            num = sum;
        }
        return num;
        
        //Method 2
        if(num == 0){
            return 0;
        }
        if(num%9 == 0){
            return 9;
        }
        return num%9;
        */
        
        //Method 3
        if(num == 0){
            return 0;
        }
        return (num-1)%9+1;
    }
}