Program for print reverse string of given number in java?

Answer 1

public class StringReverseExample

{

public static void main(String[] args)

{

int num=1001;

int n=num, rev;

while(num!=0)

{

int d=num%10;

rev= (rev*10)+d;

num=num/10;

}

System.uot.println(rev);

}

}

Answer 2

To reverse the digits, you mean? I would convert to a StringBuffer, and use the built-in reverse() method to reverse it.

If you want to program this yourself, because that is what a teacher wants, get the last digit and print it out (or add it to a string); then divide the number by 10, and again get the last digit of the result (which is the second-last digit from the original number), and continue until you have naught left. To get the last digit, use the "%" operator:

myNumber % 10

This will give you the remainder of the division by 10.

To reverse the digits, you mean? I would convert to a StringBuffer, and use the built-in reverse() method to reverse it.

If you want to program this yourself, because that is what a teacher wants, get the last digit and print it out (or add it to a string); then divide the number by 10, and again get the last digit of the result (which is the second-last digit from the original number), and continue until you have naught left. To get the last digit, use the "%" operator:

myNumber % 10

This will give you the remainder of the division by 10.

To reverse the digits, you mean? I would convert to a StringBuffer, and use the built-in reverse() method to reverse it.

If you want to program this yourself, because that is what a teacher wants, get the last digit and print it out (or add it to a string); then divide the number by 10, and again get the last digit of the result (which is the second-last digit from the original number), and continue until you have naught left. To get the last digit, use the "%" operator:

myNumber % 10

This will give you the remainder of the division by 10.

To reverse the digits, you mean? I would convert to a StringBuffer, and use the built-in reverse() method to reverse it.

If you want to program this yourself, because that is what a teacher wants, get the last digit and print it out (or add it to a string); then divide the number by 10, and again get the last digit of the result (which is the second-last digit from the original number), and continue until you have naught left. To get the last digit, use the "%" operator:

myNumber % 10

This will give you the remainder of the division by 10.

Answer 3

class Reverse

{

public static void printReverse(int n)

{

int temp = n;

int rev = 0, b;

while ( n > 0 )

{

b = n % 10;

rev = rev * 10 + b;

n = n / 10;

}

System.out.println("The number is..." + temp);

System.out.println("The reverse of the number is..." + rev);

}

}

Answer 4

class Reverse{

public static void main(String args[]){

int num=0,rem=0,rev=0;

Scanner sc=new Scanner(System.in);

System.out.println("ENTER THE NUMBER TO BE REVERSED:");

num=sc.nextInt();

do{

rem=num%10;

rev=rev+(rem*10);

num=num/10;

}while(num!=0);

System.out.println("REVERSE OF THE NUMBER IS:"+rev);

}

}

Answer 5

To reverse the digits, you mean? I would convert to a StringBuffer, and use the built-in reverse() method to reverse it.

If you want to program this yourself, because that is what a teacher wants, get the last digit and print it out (or add it to a string); then divide the number by 10, and again get the last digit of the result (which is the second-last digit from the original number), and continue until you have naught left. To get the last digit, use the "%" operator:

myNumber % 10

This will give you the remainder of the division by 10.

Answer 6

I suggest to convert the number to a String, or better directly to a StringBuffer, object. Then use the reverse() method of the StringBuffer class.