Add Digits

最新推荐文章于 2021-02-13 03:02:39 发布

mashuangwe

最新推荐文章于 2021-02-13 03:02:39 发布

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分类专栏： LeetCode 总结

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本文介绍了一种快速求解数字根的方法，即不断累加一个非负整数的所有位数直至结果仅剩一位数的过程。文章提供了两种实现方式：迭代法和数字根法，并详细解释了数字根法背后的数学原理。

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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?

Hint:

Show Hint

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

3 methods for python with explains

Iteration method

      class Solution(object):
      def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        while(num >= 10):
            temp = 0
            while(num > 0):
                temp += num % 10
                num /= 10
            num = temp
        return num

Digital Root
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

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