258. Add Digits
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?
法1:只要大于等于10就需要进行操作分离各项相加
class Solution { public: int addDigits(int num) { while (num>=10) { int sum=0; while(num>0) { sum=sum+num%10; num=num/10; } num=sum; } return num; } };
法2:
观察1到20的所有的树根:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 1
11 2
12 3
13 4
14 5
15 6
16 7
17 8
18 9
19 1
20 2
每9个一循环,所有大于9的数的树根都是对9取余,那么对于等于9的数对9取余就是0了,为了得到其本身,而且同样也要对大于9的数适用,我们就用(n-1)%9+1这个表达式来包括所有的情况,所以解法如下:
class Solution { public: int addDigits(int num) { return (num-1)%9+1; } };