leetcode 258. Add Digits——我擦，这种要你O（1）时间搞定的必然是观察规律，总结一个公式哇

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?

解法1:

class Solution(object):
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        # 1-9=1-9
        # 10=1
        # 11=2
        # 12=3 ...
        # 18=9
        # 19=>1
        # 20=>2
        # 21=>3
        # 99=>9
        # 100=>1
        # 101=>2
        # 999=>9
        def sum_digits(n):
            ans = 0
            while n:
                ans += n%10
                n /= 10
            return ans

        ans = num
        while ans > 9:
            ans = sum_digits(ans)
        return ans
        

观察发现是一个循环数组：

class Solution(object):
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        # 1-9=1-9
        # 10=1
        # 11=2
        # 12=3 ...
        # 18=9
        # 19=>1
        # 20=>2
        # 21=>3
        # 99=>9
        # 100=>1
        # 101=>2
        # 999=>9
        if num == 0: return 0
        return 9 if num % 9 == 0 else num % 9