CODEWITHSUNDEEP
java
Sandeep Johal
Sandeep Johal

Reputation: 429

How to conduct permutation using an arraylist

I am trying to compare the first element in an arraylist to the rest of the elements.

Then comparing the second element of the arraylist to the rest of the elements and so on until
the end of the arraylist.

The code is as below:

ArrayList<String> a = new ArrayList<String>();

    a.add("G1");
    a.add("G2");
    a.add("G3");
    a.add("G1");

    System.out.println(a.size());

    for(int i = 0; i<a.size(); i++){

        for(int j = 0; j<a.size(); j++){

        if(a.get(i) == a.get(j))
            System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);

        }





    }

Upvotes: 0

Views: 308

Answers (4)

Archit Saxena
Archit Saxena

Reputation: 1547

Use this code, instead of yours.

for(int i = 0; i<a.size()-1; i++){

    for(int j = i+1; j<a.size(); j++){

    if(a.get(i).equals(a.get(j)))
        System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);

    }

Hope that helps.. :)

Upvotes: 1

Smit
Smit

Reputation: 4715

Use 

 if((a.get(i)).equals(a.get(j)))

instead of 

 if(a.get(i) == a.get(j))

Morover start your initialization of j = i+1 You have already checked these previous string so no need to start it again.

EDIT

You have to limit your outer loop to i<a.size()-1 so that it wont check very last element with itself.

I hope this helps. if you need anymore help just ask.

Upvotes: 0

nhydock
nhydock

Reputation: 2416

In your case, since you're using Strings, instead of using ==, which is referential equality, as in are the memory addresses the same, you want to use .equals() which will compare the two strings by their actual values.

As for your for-loop, you can make it slightly more efficient by doing this.

for (int i = 0; i < a.size()-1; i++)
{
    for (int j = i+1; j < a.size(); j++)
    {
        if(a.get(i).equals(a.get(j)))
        {
            System.out.println("Element: " + a.get(i)+ " at " + i + " and " + "Element: "+ a.get(j)+ " at " + j);
        }

    }
}

Since there are times when you already compare a[i] with a[j], the result will be the same if you do a[j] equals a[i], so you can just skip over them. This also saves you from checking a[i] equals a[j] when j and i are the same.

Upvotes: 0

Amir Afghani
Amir Afghani

Reputation: 38511

== is reference equality (e.g do these two objects point to the same location in memory). For object equality, use .equals() instead.

Upvotes: 1

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