How to stop a while loop in correct place?

I need help because I can't stop the while loop in the correct place.
On this example, I want to remove the last char as long as the String "seq" will be the same as some number on the list (867), then stop.

More advanced is, after finding the String, to add them to new ArrayList, then try to find the rest of String (75). If found, also input to new Array.

All better solutions are very welcome. Thanks!

The ArrayList list can contain hundreds of elements, there are 4 only, some elements in the list can repeat more than one time. Find first on the list is fine for me at this moment. If found, split 86775 to 867:75. Remove 867 from String seq and add to new ArrayList. Finally, we have to find 75 - if it doesn't exist, split to 7:5, and try to find 7 and 5. On my list, single numbers always exist so no problem with exceptions.

If you want to use split method always I need to find the higher number as possible, then if not found split and removing last one but I need every digit.

The String being examined can have 20 chars or more. This is the simplest example possible.

I was looking for many methods to solve this problem and I am not sure this is best.

public class Main {
    public static void main(String[] args) {

        List<String> list = new ArrayList<String>();
        list.add("567");
        list.add("867");
        list.add("86");
        list.add("75");
        System.out.println(list);
        String seq = "86775";
        System.out.println(seq + ": found ?: " + getPatternFound(String.valueOf(list), seq) + " times");
        int nbChar = 0;
        do {
            getRemoveLastChar(seq, 1);
            if (getPatternFound(String.valueOf(list), seq) == 0) ;
            getRemoveLastChar(seq, 2);
            nbChar++;
            System.out.println(getRemoveLastChar(seq, nbChar));

        }
        while (nbChar < seq.length());

    }

    private static String getRemoveLastChar(String str, int nbChar) {
        return str.substring(0, str.length() - nbChar);
    }

    private static int getPatternFound(String longString, String pat) {
        Pattern pattern = Pattern.compile(pat);
        Matcher matcher = pattern.matcher(longString);
        int count = 0;
        while (matcher.find())
            count++;
        return count;
    }
}

[567, 867, 86, 30]
867755: found ?: 0 times
86775
8677
867 <- stop here
86
8

To answer the main question: how do you quit when done?

Of course you can do it in-loop with the break or continue keywords.
What I deem more elegant is the use of return in a separate method.
It makes the code easier to understand if later you have changes.

Here is an implementation that shows the behaviour as you stated:

import java.util.ArrayList;
public class Test {
public static void main(String[] args) {
ArrayList<String> list = new ArrayList<String>();
list.add("567");
list.add("867");
list.add("86");
list.add("75");
System.out.println("List = " + list);
String seq = "86775";
String result = search(list, seq);
}
/**
* 
* @param list
* @param seq
* @return null means no match;
*/
private static String search(ArrayList<String> list, String seq) {
while (seq.length() > 0) {
System.out.println("analyzing seq: " + seq);
boolean broken = false;
for (String s : list) {
if (seq.contentEquals(s)) { // if matched
System.out.println("found match: " + s);
// do your logic, for example
seq = seq.replace(s, "");
// quit the loop at right place
return s;
}
}
seq = seq.substring(0, seq.length() - 1);
}
return null;
}
}

generates output:

List = [567, 867, 86, 75]
analyzing seq: 86775
analyzing seq: 8677
analyzing seq: 867
found match: 867

When reading what you want to do with it: the String API offers String.contains(String infix) in combination with String.replace(String old, String new) . These should solve your real problem much better than doing everything manually in loops.