Time comparison

Java doesn't (yet) have a good built-in Time class (it has one for JDBC queries, but that's not what you want).

One option would be use the JodaTime APIs and its LocalTime class.

Sticking with just the built-in Java APIs, you are stuck with java.util.Date. You can use a SimpleDateFormat to parse the time, then the Date comparison functions to see if it is before or after some other time:

SimpleDateFormat parser = new SimpleDateFormat("HH:mm");
Date ten = parser.parse("10:00");
Date eighteen = parser.parse("18:00");

try {
    Date userDate = parser.parse(someOtherDate);
    if (userDate.after(ten) && userDate.before(eighteen)) {
        ...
    }
} catch (ParseException e) {
    // Invalid date was entered
}

Or you could just use some string manipulations, perhaps a regular expression to extract just the hour and the minute portions, convert them to numbers and do a numerical comparison:

Pattern p = Pattern.compile("(\d{2}):(\d{2})");
Matcher m = p.matcher(userString);
if (m.matches() ) {
    String hourString = m.group(1);
    String minuteString = m.group(2);
    int hour = Integer.parseInt(hourString);
    int minute = Integer.parseInt(minuteString);

    if (hour >= 10 && hour <= 18) {
        ...
    }
}

It really all depends on what you are trying to accomplish.

example:

import java.util.*;   
import java.lang.Object;   
import java.text.Collator;   
public class CurrentTime{   
  public class CurrentTime   
{   
    public static void main( String[] args )   
    {   
        Calendar calendar = new GregorianCalendar();   
        String am_pm;   
        int hour = calendar.get( Calendar.HOUR );   
        int minute = calendar.get( Calendar.MINUTE );   
        // int second = calendar.get(Calendar.SECOND);   
        if( calendar.get( Calendar.AM_PM ) == 0 ){   
            am_pm = "AM";   
            if(hour >=10)   
                System.out.println( "welcome" );   
        }               
        else{   
            am_pm = "PM";   
            if(hour<6)   
                System.out.println( "welcome" );   
        }   

        String time = "Current Time : " + hour + ":" + minute + " " + am_pm;   
        System.out.println( time );    
    }   
}  

Source

From your statement, it seems like you merely want to write:

if (10 >= hh && hh < 18) {
  ...
}

This is trivial if you are given the hours already. But surely you are asking something else?

The following assumes that your hours and minutes are stored as ints in variables named hh and mm respectively.

if ((hh > START_HOUR || (hh == START_HOUR && mm >= START_MINUTE)) &&
        (hh < END_HOUR || (hh == END_HOUR && mm <= END_MINUTE))) {
    ...
}

        String timeRange = "12:24-13:24";
        String[] timeR = timeRange.trim().split("-");
        
        // java 8
        LocalTime start = LocalTime.parse(timeR[0].trim());
        LocalTime end = LocalTime.parse(timeR[1].trim());
        LocalTime current = LocalTime.now();
        LocalTime currentHM = LocalTime.parse(current.getHour()+":"+current.getMinute());
        
        
        if(!currentHM.isBefore(start) && !currentHM.isAfter(end)) {
            return true;
        }else {
            return false;
        }

You can use the compareTo() method from Java Date class

public int result = date.compareTo(Date anotherDate); 

Return Value: The function gives three return values specified below:

It returns the value 0 if the argument Date is equal to this Date. It returns a value less than 0 if this Date is before the Date argument. It returns a value greater than 0 if this Date is after the Date argument.

I don' t know Java, but in Kotlin we have coerceIn now.

public fun <T : Comparable<T>> T.coerceIn(
    range: ClosedRange<T>
): T

Ensures that this value lies in the specified range.

Returns: this value if it's in the range, or range.start if this value is less than range.start, or range.endInclusive if this value is greater than range.endInclusive.