Python实现"各位相加"的两种方法

给定一个非负整数num,重复累加num中的每一个数字直至结果为个位数

Input: 38
Output: 2 
Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
Since 2 has only one digit, return it

进阶:

不用循环或者递归,并且时间复杂度为O(1)

1:不停循环累加,直至结果小于10

def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        while num>9:
            newNum = 0
            for i in str(num):
                newNum += int(i)
            num = newNum
        return num

2:(13)%9=(1*10+3)%9=(1*9+1+3)%9=(1+3)%9 故:(ab)%9=(a+b)%9(参考他人代码)

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

算法题来自:https://leetcode-cn.com/problems/add-digits/description/

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转载自blog.csdn.net/qiubingcsdn/article/details/82826770