Compare strings in two different arraylist (JAVA)

Question

This is a pice of my code :

  ArrayList<String> Alist= new ArrayList<String>();
  ArrayList<String> Blist= new ArrayList<String>(); 

  Alist.add("gsm");
  Alist.add("tablet");
  Alist.add("pc");
  Alist.add("mouse");

  Blist.add("gsm");
  Blist.add("something");
  Blist.add("pc");
  Blist.add("something");

so i have two array list i want to compare all items and check if they are not equal and if they are to print out only the items that are not equal.

so i make something like this:

http://postimage.org/image/adxix2i13/ sorry for the image but i have somekind of bug when i post here a for looop.

and the result is :

not equals..:tablet
not equals..:pc
not equals..:mouse
not equals..:gsm
not equals..:tablet
not equals..:pc
not equals..:mouse
not equals..:gsm
not equals..:tablet
not equals..:pc
not equals..:mouse
not equals..:gsm
not equals..:tablet

i want to print only the 2 that are not equal in the example they are gsm and pc

not equals..:gsm
not equals..:pc

Well I can't see the image. And what bug can you get while posting a code? I think you should try posting it again. — Rohit Jain
– Rohit Jain, Commented Nov 3, 2012 at 8:58
I don't get why "gsm" and "pc" are the items "that are not equal". — Denys Séguret
– Denys Séguret, Commented Nov 3, 2012 at 9:08

Denys Séguret · Accepted Answer · 2014-12-05 10:25:22Z

11

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));
    }
}

edited Dec 5, 2014 at 10:25

answered Nov 3, 2012 at 9:01

Denys Séguret

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1 Comment

user1792440 Over a year ago

Thanks :) but this check only the one by one how can i make it many to many check

Frank · Accepted Answer · 2012-11-03 09:03:48Z

1

You can use the RemoveAll(Collection c) on one of the lists, if you happen to know if one list always contains them all.

edited Nov 3, 2012 at 9:03

answered Nov 3, 2012 at 9:00

Frank

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Comments

Srinivas · Accepted Answer · 2012-11-03 09:06:04Z

0

You could use the following code:

    ArrayList<String> Alist = new ArrayList<String>();
    ArrayList<String> Blist = new ArrayList<String>();

    Alist.add("gsm");
    Alist.add("tablet");
    Alist.add("pc");
    Alist.add("mouse");

    Blist.add("gsm");
    Blist.add("something");
    Blist.add("pc");
    Blist.add("something");
    for (String a : Alist)
    {
        for (String b : Blist)
        {
            if (a.equals(b))
            {
                System.out.println("Equals " + a);
                break;
            }
        }
    }

Output is:
Equals gsm
Equals pc

answered Nov 3, 2012 at 9:06

Srinivas

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2 Comments

user1792440 Over a year ago

Ok but if i want ot check the first item of Alist with the last item or Blist

Srinivas Over a year ago

@user1792440 I don't seem to understand your query above. Can you explain by modifying your example?

jlordo · Accepted Answer · 2012-11-03 09:06:46Z

0

right now your comparing each element to all of the other ones. Do something like

for (int i = 0; i < Alist.size(); i++) {
    if (!Alist.get(i).equals(Blist.get(i)) {
        // print what you want
    }
}

Thats of course assuming both lists have the same length.

answered Nov 3, 2012 at 9:06

jlordo

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3 Comments

user1792440 Over a year ago

Ok but we dont know that every time the item will be same on the same line how can i check many to many

jlordo Over a year ago

Sort AList and BList before checking? But then you have a problem if the corresponding items are not aligned.

jlordo Over a year ago

give me a minute, i'll come up with a clever new answer

ct_ · Accepted Answer · 2012-11-03 09:22:43Z

0

Rather than writing code to manually compare list elements you might consider using Apache Commons Collections.

import org.apache.commons.collections.CollectionUtils;

List listA = ...;
List listB = ...;

Collection intersection = CollectionUtils.intersection(listA, listB);

answered Nov 3, 2012 at 9:22

ct_

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1 Comment

ct_ Over a year ago

Check the hashCode and equals methods of the objects in the lists.

jlordo · Accepted Answer · 2012-11-03 09:35:40Z

0

import java.util.HashSet;

public class CheckSet<T> extends HashSet<T>{

    @Override
    public boolean add(T e) {
        if (contains(e)) {
            remove(e);
            return true;
        } else {
            return super.add(e);
        }
    }   
}

Add all elements of both of your lists to a CheckSet intance, and at the end it will only contain the ones not equal.

answered Nov 3, 2012 at 9:35

jlordo

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Comments

stones333 · Accepted Answer · 2013-04-09 01:56:41Z

0

Here is one way:

    public static boolean compare(List<String> first, List<String> second) {
    if (first==null && second==null) return true;
    if (first!=null && second==null) return false;
    if (first==null && second!=null) return false;

    if ( first.size()!=second.size() ) return false;

    HashMap<String, String> map = new HashMap<String, String>();
    for (String str : first) {
        map.put(str, str);
    }
    for (String str : second) {
        if ( ! map.containsKey(str) ) {
            return false;
        }
    }
    return true;
}

answered Apr 9, 2013 at 1:56

stones333

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Comments

Raj · Accepted Answer · 2016-02-18 10:21:10Z

0

public static void main(String args[] ) throws Exception {

     List<String> arrayList1 = new ArrayList<String>();
     arrayList1.add("a");
     arrayList1.add("b");
     arrayList1.add("c");
     arrayList1.add("d");

     List<String> arrayList2 = new ArrayList<String>();
     arrayList2.add("a");
     arrayList2.add("b");
     arrayList2.add("c");
     arrayList2.add("d");

     boolean isEqual = false;
     if(arrayList1.size() == arrayList2.size()){
        List<String> arrayListTemp = new ArrayList<String>();
        arrayListTemp.addAll(arrayList1);
        arrayListTemp.addAll(arrayList2);
        HashSet<Object> hashSet = new HashSet<Object>();
        hashSet.addAll(arrayListTemp);
        if(hashSet.size() == arrayList1.size() &&
            hashSet.size() == arrayList2.size()){
            isEqual = true;
        }
     }
     System.out.println(isEqual);
 }

answered Feb 18, 2016 at 10:21

Raj

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1 Comment

Uwe Allner Over a year ago

PLease provide some context for your answer. See here: stackoverflow.com/help/how-to-answer

Deepak gupta · Accepted Answer · 2019-06-28 09:03:46Z

0

we can compare two different size arrayList in java or Android as follow.

ArrayList<String> array1 = new ArrayList<String>();
ArrayList<String> array2 = new ArrayList<String>();

array1.add("1");
array1.add("2");
array1.add("3");
array1.add("4");
array1.add("5");
array1.add("6");
array1.add("7");
array1.add("8");

array2.add("1");
array2.add("2");
array2.add("3");
array2.add("4");

 for (int i = 0; i < array1.size(); i++) {
            for (int j=0;j<array2.size();j++) {
                if (array1.get(i) == array2.get(j)) {
                    //if match do the needful
                } else {
                    // if not match
                }
            }
        }

answered Jun 28, 2019 at 9:03

Deepak gupta

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Comments

user3415060 · Accepted Answer · 2014-03-13 11:16:58Z

-2

import java.util.Arrays;
public class ExampleContains {
    public static boolean EligibleState(String state){
        String[] cities = new String[]{"Washington", "London", "Paris", "NewYork"};
        boolean test = Arrays.asList(cities).contains(state)?true:false;
        return test;
     }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println(EligibleState("London"));
    }

}

answered Mar 13, 2014 at 11:16

user3415060

1

1 Comment

FazoM Over a year ago

It doesn't compare 2 Lists.

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