This is an homework problem
Is there a way tor reverse a number in Java without using any loops? The only solution I can think of is reversing it using String and then casting it back to an integer.
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Your question has the answer. What is it the you want?Pramesh Bajracharya– Pramesh Bajracharya2018-11-18 05:00:04 +00:00Commented Nov 18, 2018 at 5:00
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2Under the covers converting a number to a string, reversing a string and converting a string to a number all have loops under the covers. In fact, just about any sensible program to do this entails looping, one way or another. You need to tell us the requirements precisely as they were given to you.Stephen C– Stephen C2018-11-18 06:36:32 +00:00Commented Nov 18, 2018 at 6:36
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.Homework problems expect you to demonstrate a technique so even if you could solve the problem another way, you have to solve it the expected way to get full marks.Peter Lawrey– Peter Lawrey2018-11-18 12:55:01 +00:00Commented Nov 18, 2018 at 12:55
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7 Answers
If you want to reverse a number withour using any loop you can use Recursion method call. Following program is doing same
public static void reverseMethod(int number) {
if (number < 10) {
System.out.println(number);
return;
} else {
System.out.print(number % 10);
reverseMethod(number / 10);
}
}
public static void main(String args[]) {
int num = 4567;
reverseMethod(num);
}
1 Comment
huseyin tugrul buyukisik
What if jvm does tail-recursion optimization and converts to loop?
Even if you were to reverse the number by casting it into a String, you would still need a loop if you want the program to work when having ints of different sizes. If I were to make a method to reverse a number but could not do it with loops, I would probably do it with recursion (which still uses loops indirectly). The code will look something like this:
class Main {
public static void main(String[] args) {
String input = "1234"; // or scanner to take in input can be implemented
System.out.println(Integer.parseInt(reverseInt(input)));
}
public static String reverseInt(String x) {
if (x.length() == 1) {
return x;
} else {
return x.substring(x.length() - 1) + reverseInt(x.substring(0, x.length() - 1));
}
}
}
Hope this helps!
Comments
By using reverse() of StringBuilder:
int number = 1234;
String str = String.valueOf(number);
StringBuilder builder = new StringBuilder(str);
builder.reverse();
number = Integer.parseInt(builder.toString());
System.out.println(number);
will print:
4321
Comments
if you want reverse method without loop and recursion then use this code
int a=12345;
int b,c,d,e,f;
b=a%10;
c=a%100/10;
d=a%1000/100;
e=a%10000/1000;
f=a%100000/10000;
System.out.println(b+","+c+","+d+","+e+","+f);
Comments
you can go like :
public int reverse(int x) {
String o = "";
if (x < 0) {
x *= -1;
String s = Integer.toString(x);
o += "-";
o += new StringBuilder(s).reverse().toString();
}
else {
String s = Integer.toString(x);
o += new StringBuilder(s).reverse().toString();
}
try {
int out = Integer.parseInt(o);
//System.out.println(s);
return out; }
catch (NumberFormatException e) {
return 0;
}
}
Comments
This is a solution using recursive method call
public class Tester{
public static int findReverse(int num, int temp){
if(num==0){
return temp;
}else if(num<10){
return temp*10 + num; //up to this is stopping condition
}else{
temp = temp*10 + num%10;
return findReverse(num/10, temp);
}
}
public static void main(String args[]){
int num = 120021;
int reverseNum = findReverse(num, 0);
System.out.println(reverseNum);
if(num == reverseNum)
System.out.println(num +" is a palindrome!");
else
System.out.println(num +" is not a palindrome!");
}
}
This will be fast.
static int reverseNum(int num) {
StringBuilder sb = new StringBuilder(String.valueOf(num));
sb.reverse();
return Integer.parseInt(sb.toString());
}
