[LeetCode] 258. Add Digits_Easy tag: Math
2018-08-11 11:59 Johnson_强生仔仔 阅读(158) 评论(0) 编辑 收藏 举报Given a non-negative integer num
, repeatedly add all its digits until the result has only one digit.
Example:
Input:38
Output: 2 Explanation: The process is like:3 + 8 = 11
,1 + 1 = 2
. Since2
has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
基本方法, 用一个helper function, 去计算num的digits sum, 然后如果不是< 10, 继续循环call helper function.
Math 方法: O(1)
The problem, widely known as digit root problem, has a congruence formula:
https://en.wikipedia.org/wiki/Digital_root#Congruence_formula
For base b (decimal case b = 10), the digit root of an integer is:
- dr(n) = 0 if n == 0
- dr(n) = (b-1) if n != 0 and n % (b-1) == 0
- dr(n) = n mod (b-1) if n % (b-1) != 0
or
- dr(n) = 1 + (n - 1) % 9
Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).
Code
1) while
class Solution: def addDigits(self, num): """ :type num: int :rtype: int """ def helper(num): return sum(int(c) for c in str(num)) while num > 9: num = helper(num) return num
2) Math
class Solution: def addDigits(self, num): if num == 0: return 0 return 1 + (num-1)%9