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 HintA 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 HintHow do they occur, periodically or randomly?Show More HintYou may find this Wikipedia article useful.3 methods for python with explainsIteration methodclass 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 numDigital Rootthis 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 = 110 % 9 = 1100 % 9 = 1so N % 9 = a[0] + a[1] + ..a[n]means N % 9 = Mso 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 = 9class Solution(object):def addDigits(self, num): """ :type num: int :rtype: int """ if num == 0 : return 0 else:return (num - 1) % 9 + 1
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