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. Since 2 has only one digit, return it. Follow up: Could you do it without any loop/recursion in O(1) runtime?
public class Solution {
public int AddDigits(int num) {
if(num<10){
return num;
}
int sum = 0;
while(num>0){
sum+=num%10;
num/=10;
}
return AddDigits(sum);
}
}Time Complexity: O(n)
Space Complexity: O(n)
Efficient Approach:
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).
From the formula, we can find that the result of this problem is immanently periodic, with a period (b-1).
Output sequence for decimals (b = 10):
~input: 0 1 2 3 4 ... output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....
Henceforth, we can write the following code, whose time and space complexities are both O(1).
public class Solution {
public int AddDigits(int num) {
return 1 + (num-1)%9;
}
}
