CODEWITHSUNDEEP
javacollectionsdiffcomparesequence
Abhishek
Abhishek

Reputation: 253

Sequence comparison in java

I'm looking for a standard algorithm/code (Java) which compares two integer lists (old and new) and gives a third result list which provides actions to convert the 'old' list into the 'new' list.

For example:

old-> 1, 2, 3, 4
new-> 9, 2, 3, 6, 4

so the result should be something like:

1-, 9+, 2, 3, 4-, 6+, 4+

Here, the suffix:

  - = Deleted item from old list.
  + = New added item to old list.

and the rest (w/o suffix), are numbers which are unchanged (i.e Value as well as Index). I believe something using the LCS (longest common sequence) would do this job! But I can't really figure-out if there is any.

Any pointers will be highly appreciated.

Upvotes: 2

Views: 2372

Answers (3)

Abhishek
Abhishek

Reputation: 253

Just for future references. Both the 1st and 2nd answers are good. The 1st answer is clue to what i was looking for. The optimal way to compare sequences. and, The 2nd answer is a working code to compare sequences. But this doesn't give a optimal result for coverting one list to another. But good for a simple diff!!

Thanks all for the answers!!

Thanks, Abhishek.

Upvotes: 0

bruno conde
bruno conde

Reputation: 48265

I implemented something in C#. Porting it to Java ...

(edit)

Here is the Java version:

enum Action {
    UNCHANGED, ADDED, REMOVED
}

static class DiffResult<T> {
    private T value;
    public Action type;

    public DiffResult(T value, Action type) {
        super();
        this.value = value;
        this.type = type;
    }

    public T getValue() {
        return value;
    }

    public Action getType() {
        return type;
    }
}


public static <T> List<DiffResult<T>> listDiff(List<T> originalList,
        List<T> newList) {
    List<DiffResult<T>> result = new ArrayList<DiffResult<T>>();

    int maxCount = Math.max(originalList.size(), newList.size());
    for (int i = 0; i < maxCount; i++) {
        if (newList.size() < i + 1)
            result.add(new DiffResult<T>(originalList.get(i),
                    Action.REMOVED));
        else {
            if (originalList.size() < i + 1) {
                result.add(new DiffResult<T>(newList.get(i), Action.ADDED));
            } else {
                if (originalList.get(i).equals(newList.get(i)))
                    result.add(new DiffResult<T>(originalList.get(i),
                            Action.UNCHANGED));
                else {
                    result.add(new DiffResult<T>(originalList.get(i),
                            Action.REMOVED));
                    result.add(new DiffResult<T>(newList.get(i),
                            Action.ADDED));
                }
            }
        }
    }
    return result;
}

public static void main(String[] args) {
    List<Integer> oldList = new ArrayList<Integer>();
    oldList.add(1);
    oldList.add(2);
    oldList.add(3);
    oldList.add(4);

    List<Integer> newList = new ArrayList<Integer>();
    newList.add(9);
    newList.add(2);
    newList.add(3);
    newList.add(6);
    newList.add(4);

    List<DiffResult<Integer>> diff = listDiff(oldList, newList);

    for (DiffResult<Integer> d : diff) {
        System.out.println("Item: " + d.getValue() + " -> " + d.getType());
    }
}

Upvotes: 2

Mehrdad Afshari
Mehrdad Afshari

Reputation: 421978

Levenshtein distance algorithm seems to work for you (essentially the LCS algorithm you mentioned). Just record the action you choose in another table (right after when you choose the minimum, you need to record which action has resulted the min cost to be able to look it up afterward).

if (seq1[i] == seq2[j] && d[i - 1, j - 1] <= d[i - 1, j] + 1
                       && d[i - 1, j - 1] <= d[i, j - 1] + 1) {
     d[i, j] = d[i - 1, j - 1];
     action[i, j] = MATCHED;
} else if (d[i - 1, j] < d[i, j - 1]) // If cost of insertion is less:
{
     d[i, j] = d[i - 1, j] + 1;
     action[i, j] = INSERTION;
} else {
     d[i, j] = d[i, j - 1] + 1;
     action[i, j] = DELETION;
}

Then use action[i, j] to recursively go back through the process and pushing the chosen action in a stack.

Upvotes: 3

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