258. Add Digits

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本文介绍LeetCode上258题Add Digits的解题思路与两种实现方法。一种采用递归方式，另一种则利用数学规律，实现O(1)时间复杂度的解决方案。

258. Add Digits

Leetcode link for this question

Discription:

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?

Analyze:

Digital root

Code 1:

class Solution(object):    
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        if (num==0):
            return 0
        else:
            return 1+(num-1)%9

Submission Result:

Status: Accepted
Runtime: 56 ms
Ranking: beats 95.55%

Code 2:

class Solution(object):
    def addDigits1(self, num):
        """
        :type num: int
        :rtype: int
        """
        s='%d' %num
        c=0
        for i in range(len(s)):
            c=c+int(s[i])
        if(c<10):
            return c
        else:
            return self.addDigits(c)