Java Count Number of Digits in a given Number
I have given here Java code to count the number of digits in a given number. It counts the digits using division operation in do while loop. The alternate way to count the number of digits is using log (base 10) operation.
Source Code
import java.lang.*;
import java.util.*;
import java.io.*;
class ReverseNumber
{
public static String ReadString()
{
try
{
InputStreamReader input = new InputStreamReader(System.in);
BufferedReader reader = new BufferedReader(input);
return reader.readLine();
}
catch (Exception e)
{
e.printStackTrace();
return "";
}
}
public static int ReadInteger()
{
try
{
InputStreamReader input = new InputStreamReader(System.in);
BufferedReader reader = new BufferedReader(input);
return Integer.parseInt(reader.readLine());
}
catch (Exception e)
{
e.printStackTrace();
return 0;
}
}
static int CountNumberOfDigits_UsingLog(int number)
{
return (int)(Math.log((double)number) + 1);
}
static int CountNumberOfDigits(int number)
{
int numdgits = 0;
do
{
number = number / 10;
numdgits++;
} while (number > 0);
return numdgits;
}
public static void main(String[] args)
{
System.out.println("\nProgram to find the reverse of a number. Enter -1 to exit");
while(true)
{
System.out.print("\n\nEnter a Number (-1 to exit): ");
int n = ReadInteger();
if(n < 0)
break;
int number = n;
int index = 0;
int numdigits = CountNumberOfDigits(number);
System.out.format("Number of digits in %d is: %d\n", number, numdigits);
int result = 0;
for (int i = 0; i < numdigits; i++)
{
result *= 10;
result += n % 10;
n = n / 10;
}
System.out.format("The reverse of number %d is: %d\n", number, result);
}
}
}
Output
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