LeetCode 258. Add Digits(数字相加)
题目描述
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?
解法一
思路
这题没啥意思,算是一个经典的数学问题,叫做Digital Root,其解就是f(n) = 1 + (n - 1) % 9
。
Python
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
if num == 0:
return 0
return 1 + (num - 1) % 9
Java
public class Solution {
public int addDigits(int num) {
return 1 + (num - 1) % 9;
}
}
C++
class Solution {
public:
int addDigits(int num) {
return 1 + (num - 1) % 9;
}
};