Question

Formatted question description: https://leetcode.ca/all/258.html

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38
Output: 2 
Explanation: 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:
  A naive implementation of the above process is trivial. Could you come up with other methods?
  What are all the possible results?
  How do they occur, periodically or randomly?
  You may find this Wikipedia article useful.

Algorithm

Let’s first observe all the roots from 1 to 20:

1    1
2    2
3    3
4    4
5    5
6    6
7    7
8    8    
9    9    
10    1
11    2
12    3    
13    4
14    5
15    6
16    7
17    8
18    9
19    1
20    2

Draw a rule, every 9 a cycle, all roots of numbers greater than 9 take the remainder of 9, so for the number equal to 9 to take the remainder of 9 is 0, in order to get itself, and also need to be greater than The number of 9 applies, we use the expression (n-1)%9+1 to cover all cases.

There is another special case that needs to be considered. When num is 0, then there will be a situation of -1% 9, need to make an additional judgment.

Code

Java

  • class Solution {
      public int addDigits(int num) {
              return (num == 0) ? 0 : (num - 1) % 9 + 1;
          }
    };
    
    
  • // OJ: https://leetcode.com/problems/add-digits/
    // Time: O(lgN)
    // Space: O(1)
    class Solution {
    public:
        int addDigits(int n) {
            while (n > 9) {
                int next = 0;
                for (; n; n /= 10) next += n % 10;
                n = next;
            }
            return n;
        }
    };
    
  • class Solution(object):
      def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        if num < 10:
          return num
        return 1 + (num - 1) % 9
    
    

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