leetcode 258 Add Digits

简单来说就是求给定一个数的每个位置数值的相加

#coding=utf-8

'''

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?

'''

class Solution(object):

    def addDigits1(self, num):

        """

        :type num: int

        :rtype: int

        """

        if not num:return

        while num>=10:

            num=sum([int(str(x)) for x in str(num)])

        return num

    #下面两个是 @xili_keke 的代码

    #掌握这个方法的思路，应该很常用

    def addDigits2(self,num):

        while num>=10:

            temp=0

            while num>0:

                temp+=num%10

                num/=10

            num=temp

        return num

    

def addDigits3(self,num):

        """

            :type num: int

            :rtype: int

            """

        if num == 0:

            return 0

        else:

            return (num - 1) % 9 + 1

        '''this method depends on the truth:

N=(a[0] * 1 + a[1] * 10 + ...a[n] * 10 ^n),and a[0]...a[n] are all between [0,9]

we set M = a[0] + a[1] + ..a[n]

and another truth is that:

1 % 9 = 1

10 % 9 = 1

100 % 9 = 1

so N % 9 = a[0] + a[1] + ..a[n]

means N % 9 = M

so N = M (% 9)

as 9 % 9 = 0,so we can make (n - 1) % 9 + 1 to help us solve the problem when n is 9.as N is 9, ( 9 - 1) % 9 + 1 = 9'''

s=Solution()

print s.addDigits2(12123)