258. Add Digits

258. Add Digits

Total Accepted: 79895 Total Submissions: 165648 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?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

一般的想法就是把每个数的个位，十位都分开，再一次次算：

    class Solution {  
    public:  
        int addDigits(int num) {  
            int i=0, x=2;  
            while(x>1) {  
                i=0;
                x=0;  
                while(num) {  
                    i+=num%10;
                    num=num/10;  
                    x++;  
                }  
                if(x == 1)  
                    break;  
                num =i;
            }  
              
            return i;  
        }  
    };  

还可用递归：

class Solution {
public:
    int addDigits(int num) {
    int sum=0;    
        while(num){    
            sum=sum+num%10;    
            num=num/10;    
        }    
        if(sum<10){    
            return sum;    
        }    
        else{    
            return addDigits(sum);    
        }    
}
};