Problem: Write a Java Program to convert a binary number into its corresponding decimal representation.

Example:

```Input: 101
Output: 5

Input: 1111
Output: 15```

## Method 1: Using While Loop

To convert a binary into a decimal we multiply the digits of the binary with their respective exponential value of 2 (2position from right – 1) and sum them up.

To iterate through the digits in Java, we use the while loop as follows:

``````import java.util.Scanner;

public class Main
{
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int lastDigit, decimal=0, i=0;

//Input binary number
System.out.println("Enter a Binary");
int binary = in.nextInt();

while(binary>0){
//extract digit
lastDigit = binary%10;

//multiply digit with the 2^i and add to the sum
decimal += Math.pow(2,i) * lastDigit;

//update binary by removing the last digit
binary = binary/10;

i++;    //Increment exponent ('i')
}

//output decimal
System.out.println("Decimal: "+decimal);
}
}``````

Output:

Enter a Binary: 101
Decimal: 5

## Method 2: Using Recursion

The approach used in this method is the same as the above but implemented in a recursive fashion.

On every recursive call, we extract the last digit, multiply it with its corresponding exponential value of 2 and return its sum with the next recursive call result.

``````import java.util.Scanner;

public class Main
{
public static void main(String[] args) {
Scanner in=new Scanner(System.in);

//Input the binary number
System.out.println("Enter a Binary");
int binary = in.nextInt();

//pass the binary number to the recursive function
//and print the result returned
System.out.println("Decimal: "+binToDec(binary, 0));
}

//'i' represents the nth recursive call as well as
//nth digit from left, so it can be used as an exponent
private static int binToDec(int bin, int i){
//base condition
if(bin == 0 || bin == 1)
return (int)Math.pow(2,i)*bin;

//extract digit
int lastDigit = bin%10;

//return the arithemetic result with next recusive call
return (int)(Math.pow(2,i) * lastDigit) + binToDec(bin/10, ++i);

}
}``````

Output:

Enter a Binary: 1111
Decimal: 15

It is important to note that on every next recursive call, we are eliminating the last digit from the binary number that has been processed in the current recursion.

If you have any doubts or suggestions then below.