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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);

}

}

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12y ago
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14y ago

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.

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Wiki User

14y ago

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);

}

}

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Wiki User

13y ago

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);

}

}

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Wiki User

14y ago

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.

This answer is:
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Wiki User

11y ago

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.


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Q: Program for print reverse string of given number in java?
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