Java Program for Sum the digits of a given number
Last Updated :
10 Feb, 2023
Given a number, find the sum of its digits.
Example :
Input : n = 687
Output : 21
Input : n = 12
Output : 3
1. Iterative:
Java
import java.io.*;
class GFG {
static int getSum( int n)
{
int sum = 0 ;
while (n != 0 )
{
sum = sum + n % 10 ;
n = n/ 10 ;
}
return sum;
}
public static void main(String[] args)
{
int n = 687 ;
System.out.println(getSum(n));
}
}
|
C++
#include <iostream>
#include <string>
#include <vector>
class GFG
{
public :
static int getSum( int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
static void main(std::vector<std::string> &args)
{
int n = 687;
std::cout << GFG::getSum(n) << std::endl;
}
};
int main( int argc, char **argv){
std::vector<std::string> parameter(argv + 1, argv + argc);
GFG::main(parameter);
return 0;
};
|
C#
using System;
public class GFG
{
public static int getSum( int n)
{
var sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = ( int )(n / 10);
}
return sum;
}
public static void Main(String[] args)
{
var n = 687;
Console.WriteLine(GFG.getSum(n));
}
}
|
Time Complexity: O(|n|)
Auxiliary Space: O(1)
How to compute in single line?
Java
import java.io.*;
class GFG {
static int getSum( int n)
{
int sum;
for (sum = 0 ; n > 0 ; sum += n % 10 ,
n /= 10 );
return sum;
}
public static void main(String[] args)
{
int n = 687 ;
System.out.println(getSum(n));
}
}
|
Time Complexity: O(|n|)
Auxiliary Space: O(1)
2. Recursive
Java
import java.io.*;
class GFG {
static int sumDigits( int no)
{
return no == 0 ? 0 : no% 10 +
sumDigits(no/ 10 ) ;
}
public static void main(String[] args)
{
int n = 687 ;
System.out.println(sumDigits(n));
}
}
|
Time Complexity: O(log10n)
Auxiliary Space: O(log10n)
Please refer complete article on Program for Sum the digits of a given number for more details!
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