0
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);
}
}
Wiki User
∙ 11y ago
Wiki User
∙ 14y agoTo 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.
Wiki User
∙ 13y agoclass 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);
}
}
Wiki User
∙ 12y agoclass 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);
}
}
Wiki User
∙ 14y agoTo 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.
Wiki User
∙ 11y agoI 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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