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Java – Divide a series of numbers into groups of 10, 100, 1000, 10000 and so on

Specify a beginning and end of a range as a string, for example:

String start = "1230000";
String end   = "1285999";

Generate the shortest list of prefixes that span the entire range.
Thus, for each number in the specified range, there is a list item that can be used as a prefix. For the example above, the list would include:

 123,124,125,126,127,1280,1281,1282,1283,1284,1285

explanation

The length of the strings may vary (6 – 12) start.length() and end.length() are always the same.
The above range can be subdivided manually into the subregions:

    //1230000 - 1239999
    //1240000 - 1249999
    //1250000 - 1259999
    //1260000 - 1269999
    //1270000 - 1279999
    //1280000 - 1289999 this range is not completly included so make the ranges smaller with a factor 10
    //1280000 - 1280999
    //1281000 - 1281999
    //1282000 - 1282999
    //1283000 - 1283999
    //1284000 - 1284999
    //1285000 - 1285999

By storing the common substring of each start and end number of the subregions, it can be easily confirmed that any number in the range starts with 123,124,125,126,127,1280,1281,1282,1283,1284 or 1285.

Here's a very naive but working solution for printing the prefixes:

public static void main(String() args) {
    String start = "1230000";
    String end   = "1285999";

    while(start.charAt(start.length()-1) == '0' && end.charAt(start.length()-1) == '9'){
        start = start.substring(0,start.length()-1);
        end = end.substring(0,end.length()-1);
    }

    long st = Integer.parseInt(start);
    long en = Integer.parseInt(end);
    for(long i = st; i <= en; ){
         if(i % 1000 == 0 && i + 999 <= en){
            System.out.println(i/1000); 
            i = i + 1000;
        }
        else if(i % 100 == 0 && i + 99 <= en){
            System.out.println(i/100); 
            i = i + 100;
        }
        else if(i % 10 == 0 && i + 9 <= en){
            System.out.println(i/10); 
            i = i + 10;
        }
        else {
            System.out.println(i); 
            i = i + 1;
        }
    }
}

For some not so nice areas like the ones above, I have to expand my for loop:

public static void main(String() args) {
    String start = "1279995";
    String end   = "1285999";

    while(start.charAt(start.length()-1) == '0' && end.charAt(start.length()-1) == '9'){
        start = start.substring(0,start.length()-1);
        end = end.substring(0,end.length()-1);
    }

    long st = Integer.parseInt(start);
    long en = Integer.parseInt(end);
    for(long i = st; i <= en; ){
        if(i % 1000000 == 0 && i + 999999 <= en){
            System.out.println(i/1000000); 
            i = i + 1000000;
        }
        else if(i % 100000 == 0 && i + 99999 <= en){
            System.out.println(i/100000); 
            i = i + 100000;
        }
        else if(i % 10000 == 0 && i + 9999 <= en){
            System.out.println(i/10000); 
            i = i + 10000;
        }
        else if(i % 1000 == 0 && i + 999 <= en){
            System.out.println(i/1000); 
            i = i + 1000;
        }
        else if(i % 100 == 0 && i + 99 <= en){
            System.out.println(i/100); 
            i = i + 100;
        }
        else if(i % 10 == 0 && i + 9 <= en){
            System.out.println(i/10); 
            i = i + 10;
        }
        else {
            System.out.println(i); 
            i = i + 1;
        }
    }
}

Which I did not find very nice. Any ideas on how to optimize my approach or if someone has a different approach (I'm open to everything, regex, streams, recursion ...)?

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