[LeetCode] 258. Add Digits_Easy tag: Math

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?

基本方法, 用一个helper function, 去计算num的digits sum, 然后如果不是< 10, 继续循环call helper function.

Math 方法: O(1)  

The problem, widely known as digit root problem, has a congruence formula:

https://en.wikipedia.org/wiki/Digital_root#Congruence_formula

For base b (decimal case b = 10), the digit root of an integer is:

• dr(n) = 0 if n == 0
• dr(n) = (b-1) if n != 0 and n % (b-1) == 0
• dr(n) = n mod (b-1) if n % (b-1) != 0

or

• dr(n) = 1 + (n - 1) % 9

Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).

Code

1) while

class Solution:
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        def helper(num):
            return sum(int(c) for c in str(num))
        while num > 9:
            num = helper(num)
        return num

2) Math

class Solution:
    def addDigits(self, num):
        if num == 0: return 0
        return 1 + (num-1)%9