## Reverse Integer in java

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 – 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

### Input Format

The first line of input contains the integer `n`.

### Output Format

Print the integer `n` in reverse if it lies within the range, else print 0.

Input

```321
```

Output

```123
```

Input

```-321
```

Output

```-123
```

### Example 3

Input

```120
```

Output

```21
```

###Constraints

-2^31 <= x <= 2^31 – 1

## Solution of Reverse Integer in java:–

```import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc= new Scanner(System.in);
int n= sc.nextInt();
int num = n;

int curr_digit=0, rev_num=0, pre_rev_num=0;

while(num!=0){

curr_digit= num%10;
rev_num= (rev_num*10) + curr_digit;

if((rev_num-curr_digit)/10 != pre_rev_num) {
// check overflow condition
// rev_num can go out of MAX_Int range
// it will give garbage value after max range
// condition becomes true

System.out.println(0);
return;
}

pre_rev_num= rev_num;
num/=10;

}
System.out.println(rev_num);
}
}```