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?
Analyse:
1. trivival solution:
1 class Solution { 2 public: 3 int addDigits(int num) { 4 while(num >= 10) num = add(num); 5 return num; 6 } 7 int add(int num){ 8 int result = 0; 9 while(num){ 10 result += num % 10; 11 num /= 10; 12 } 13 return result; 14 } 15 };
we can find the result after implementation:
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3
4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7
8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 请按任意键继续. . .
so the result is (num - 1) % 9 + 1;
class Solution { public: int addDigits(int num) { return (num - 1) % 9 + 1; } };