Saturday, September 5, 2015

Leetcode: 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 = 111 + 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
Brute-force solution:
public class Solution {
    public int addDigits(int num) {
        if (num < 10) {
            return num;
        }
        
        int result = num;
        
        while (result >= 10) {
            // Get each didgit of the number
            int digit = 0;
            while (result > 0) {
                digit += result % 10;
                result /= 10;
            }
            
            result = digit;
        }
        
        return result;
    }
}

O(1) time solution:
Let's enumerate numbers from 1 - 19, 
in out 
1   1
2   2
3   3
 ...
9   9
10  1
11  2
12  3
13  4
14  5
...
19  1
20  2
21  3

...
28  1
29  2
30  3
...

So the result = (n - 1) % 9 + 1

Code (Java):
public class Solution {
    public int addDigits(int num) {
        if (num <= 0) {
            return 0;
        }
        
        return (num - 1) % 9 + 1;
    }
}

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