You are right in saying that if they are arrays you can't use Arrays.equals(), if they are List<>s then you cannot just use equals. In both cases they also check order.

So yes your main option without writing your own comparison algorithm is to use Collections.sort() on both lists.

If you don't need to check for duplicates you could drop both lists into a HashSet using addAll and then compare the two HashSet or just use List.containsAll. The bad news though is that these would both have the same limitation, if you need to compare lists that might have repeated elements then it may give incorrect results. i.e. "bob", "bob", "fred" would compare as equal to "bob", "fred", "fred".

Convert Lists to Collection and use removeAll

    Collection<String> listOne = new ArrayList(Arrays.asList("a","b", "c", "d", "e", "f", "g"));
    Collection<String> listTwo = new ArrayList(Arrays.asList("a","b",  "d", "e", "f", "gg", "h"));


    List<String> sourceList = new ArrayList<String>(listOne);
    List<String> destinationList = new ArrayList<String>(listTwo);


    sourceList.removeAll( listTwo );
    destinationList.removeAll( listOne );



    System.out.println( sourceList );
    System.out.println( destinationList );

Output:

[c, g]
[gg, h]

[EDIT]

other way (more clear)

  Collection<String> list = new ArrayList(Arrays.asList("a","b", "c", "d", "e", "f", "g"));

    List<String> sourceList = new ArrayList<String>(list);
    List<String> destinationList = new ArrayList<String>(list);

    list.add("boo");
    list.remove("b");

    sourceList.removeAll( list );
    list.removeAll( destinationList );


    System.out.println( sourceList );
    System.out.println( list );

Output:

[b]
[boo]

Don't use != to compare strings. Use the equals method :

if (! Blist.get(i).equals(Alist.get(j))

But this wouldn't probably fix your algorithmic problem (which isn't clear at all).

If what you want is know what items are the same at the same position, you could use a simple loop :

int sizeOfTheShortestList = Math.min(Alist.size(), Blist.size());
for (int i=0; i<sizeOfTheShortestList; i++) {
    if (Blist.get(i).equals(Alist.get(i))) {
        System.out.println("Equals..: " + Blist.get(i));
    }
}

If you want to get items that are in both lists, use

for (int i = 0; i < Alist.size(); i++) {
    if (Blist.contains(Alist.get(i))) {
        System.out.println("Equals..: " + Alist.get(i));
    }
}

You could use isEqualList from Apache's ListUtils:

isEqualList

public static boolean isEqualList(java.util.Collection list1,
                                  java.util.Collection list2)

Tests two lists for value-equality as per the equality contract in List.equals(java.lang.Object).

This method is useful for implementing List when you cannot extend AbstractList. The method takes Collection instances to enable other collection types to use the List implementation algorithm.

The relevant text (slightly paraphrased as this is a static method) is:

Compares the two list objects for equality. Returns true if and only if both lists have the same size, and all corresponding pairs of elements in the two lists are equal. (Two elements e1 and e2 are equal if (e1==null ? e2==null : e1.equals(e2)).) In other words, two lists are defined to be equal if they contain the same elements in the same order. This definition ensures that the equals method works properly across different implementations of the List interface. Note: The behaviour of this method is undefined if the lists are modified during the equals comparison.

Perhaps the easiest solution would be to use two List<List<String>>s instead. Assuming the List implementations used extend AbstractList, using equals will give you the desired behavior. From the documentation for AbstractList.equals:

Compares the specified object with this list for equality. Returns true if and only if the specified object is also a list, both lists have the same size, and all corresponding pairs of elements in the two lists are equal. (Two elements e1 and e2 are equal if (e1==null ? e2==null : e1.equals(e2)).) In other words, two lists are defined to be equal if they contain the same elements in the same order.

You can easily wrap a String[] in a thin List<String> implementation that extends AbstractList by using Arrays.asList.

EDIT: Here's an example:

String[] array1 = {"1", "2", "3"};
String[] array2 = {"4", "7"};

String[] array3 = {"1", "2", "3"};
String[] array4 = {"4", "7"};

List<List<String>> lst1 = new ArrayList<>();
lst1.add(Arrays.asList(array1));
lst1.add(Arrays.asList(array2));

List<List<String>> lst2 = new ArrayList<>();
lst2.add(Arrays.asList(array3));
lst2.add(Arrays.asList(array4));

System.out.println(lst1.equals(lst2)); //prints true

Use

ArrayList.contains("StringToBeChecked");

If the String is present in the ArrayList, this function will return true, else will return false.

Try fnlData.equals(fnlDataTMP) if both list are in order

or if order does not matter, try creating hash set and then compare using equals

new HashSet(fnlData).equals(new HashSet(fnlDataTMP))

EDIT

Here are two versions. One using ArrayList and other using HashSet

Compare them and create your own version from this, until you get what you need.

This should be enough to cover the:

P.S: It is not a school assignment :) So if you just guide me it will be enough

part of your question.

continuing with the original answer:

You may use a java.util.Collection and/or java.util.ArrayList for that.

The retainAll method does the following:

Retains only the elements in this collection that are contained in the specified collection

see this sample:

import java.util.Collection;
import java.util.ArrayList;
import java.util.Arrays;

public class Repeated {
    public static void main( String  [] args ) {
        Collection listOne = new ArrayList(Arrays.asList("milan","dingo", "elpha", "hafil", "meat", "iga", "neeta.peeta"));
        Collection listTwo = new ArrayList(Arrays.asList("hafil", "iga", "binga", "mike", "dingo"));

        listOne.retainAll( listTwo );
        System.out.println( listOne );
    }
}

EDIT

For the second part ( similar values ) you may use the removeAll method:

Removes all of this collection's elements that are also contained in the specified collection.

This second version gives you also the similar values and handles repeated ( by discarding them).

This time the Collection could be a Set instead of a List ( the difference is, the Set doesn't allow repeated values )

import java.util.Collection;
import java.util.HashSet;
import java.util.Arrays;

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

          Collection<String> listOne = Arrays.asList("milan","iga",
                                                    "dingo","iga",
                                                    "elpha","iga",
                                                    "hafil","iga",
                                                    "meat","iga", 
                                                    "neeta.peeta","iga");

          Collection<String> listTwo = Arrays.asList("hafil",
                                                     "iga",
                                                     "binga", 
                                                     "mike", 
                                                     "dingo","dingo","dingo");

          Collection<String> similar = new HashSet<String>( listOne );
          Collection<String> different = new HashSet<String>();
          different.addAll( listOne );
          different.addAll( listTwo );

          similar.retainAll( listTwo );
          different.removeAll( similar );

          System.out.printf("One:%s%nTwo:%s%nSimilar:%s%nDifferent:%s%n", listOne, listTwo, similar, different);
      }
}

Output:

$ java Repeated
One:[milan, iga, dingo, iga, elpha, iga, hafil, iga, meat, iga, neeta.peeta, iga]

Two:[hafil, iga, binga, mike, dingo, dingo, dingo]

Similar:[dingo, iga, hafil]

Different:[mike, binga, milan, meat, elpha, neeta.peeta]

If it doesn't do exactly what you need, it gives you a good start so you can handle from here.

Question for the reader: How would you include all the repeated values?

if (k == unmatchingPositions.get(k)) { This is saying if the value stored in position k of the unmatched list == k then replace it. That makes no sense. You know it's not matched so just replace it. No need for an if

You could do it all in one pass:

int listASz = listA.size();
for(int i = 0; i < listASz; i++)
{
    if (listA[i] != listB[i])
    {
        listB[i] = /*something*/

You need to extend this a bit to handle cases where the lists aren't the same size

You don't need to compare

List<Integer> c = new ArrayList<>(a);
c.removeAll(b);

And if you don't mind loosing the original list data

a.removeAll(b);