Java index of items in arrayList

Question

I'm working on a program that asks user for values, and adds them onto array list until value of -1 is inserted. Then it asks, what value do you want to search for and user inserts another value. Then the program searches for all of those values in the array list, and prints out the indexes of all the found items. My program currently prints only the first item and it's index since i'm using list.indexOf(searcheditem). How do i get it to list all of the found items and not only the 1st one? Heres my code so far.

import java.util.ArrayList;
import java.util.Scanner;

public class KysytynLuvunIndeksi {

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);

        ArrayList<Integer> list = new ArrayList<>();
        while (true) {
            int read = Integer.parseInt(reader.nextLine());
            if (read == -1) {
                break;
            }

            list.add(read);
        }

        System.out.print("What are you looking for? ");
        int searched = Integer.parseInt(reader.nextLine());

        if (list.contains(searched)) {
            System.out.println("Value " + searched + " has index of: " + (list.indexOf(searched));
        } else {
            System.out.println("value " + searched + " is not found");
        }

    }
}

Thanks in advance!

Hello and Welcome on StackOverflow ! Why not to use the same way you fill the list; but with a second list witch would contains the values searched ? — NatNgs, Feb 21 '18 at 11:08
see this [question](https://stackoverflow.com/questions/13900585/trying-to-find-all-occurrences-of-an-object-in-arraylist-in-java) — lin, Feb 21 '18 at 11:10
In your question, I happened to notice that, you mentioned "the program searches for all of those values in the array list", do you infer that your program has to take more than 1 value to search? because your program does take only 1 input. — Jayanth, Feb 21 '18 at 11:13

score 1 · Answer 1 · answered Feb 21 '18 at 11:12

1

put this for loop in if condition

if (list.contains(searched)) {
     for (int index = list.indexOf(searched);index >= 0;index =list.indexOf(searched, index + 1))
    {
      System.out.println("Value " + searched + " has index of: " + index);
    }
}

answered Feb 21 '18 at 11:12

Jay Shankar Gupta

5,918
1
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27

score 0 · Answer 2 · answered Feb 21 '18 at 11:08

0

Iterate over the list, compare items with searched, add index to the resulting list if there's a match. Something along the lines:

List<Integer> foundIndices = new ArrayList<>();
for (int index = 0; index < list.size(); index++) {
    if (list.get(index).intValue() == searched) {
        foundIndices.add(index);
    }
}

answered Feb 21 '18 at 11:08

lexicore

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Consider using a `ListIterator` for iterating over the input values. – Henrik Aasted Sørensen Feb 21 '18 at 11:14
@Henrik Why? (Honest question.) – lexicore Feb 21 '18 at 11:15
On closer inspection, I'm not sure. :) Sorry. As you were! – Henrik Aasted Sørensen Feb 21 '18 at 11:18
@Henrik Not that I actually disagree with you. It was just interesting to think about the reasons. – lexicore Feb 21 '18 at 11:20
1

Attempted a quick rewrite, and readability was not improved at all, which is often the case when people are iterating over a list and need the index. Too few people know about the `ListIterator`. :) – Henrik Aasted Sørensen Feb 21 '18 at 11:21
1

@Henrik I apreciate your effort and seriousness very much. – lexicore Feb 21 '18 at 11:23
2

`ListIterator` can (and should) be faster if list is a non-random-access type, such as a `LinkedList`. Accessing all the elements of a linked list by their indices takes O(n**2) time; a `ListIterator` does it in O(n). For `ArrayList`, as in this case, it's O(n) either way, so no great loss if not using `ListIterator`. – Kevin Anderson Feb 21 '18 at 11:25

score 0 · Answer 3 · answered Feb 21 '18 at 11:11

public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);

        ArrayList<Integer> list = new ArrayList<>();
        while (true) {
            int read = Integer.parseInt(reader.nextLine());
            if (read == -1) {
                break;
            }

            list.add(read);
        }

        System.out.print("What are you looking for? ");
        int searched = Integer.parseInt(reader.nextLine());

        for(int i=0;i<list.size();i++){
            if(list.get(i)==searched){
                System.out.println("Value " + searched + " has index of: " + (i));
            }
        }
        if (!list.contains(searched)){
            System.out.println("value " + searched + " is not found");
       }    
    }

A simple for loop would suffice.

list.contains would return the first found element in your array list.

Manos Nikolaidis · Answer 4 · 2018-02-21T11:17:29.367

A simple for loop will suffice and be faster than the combination of contains and indexOf. You may keep the loop index outside the for statement to check if nothing was found afterwards.

int i;
for (i = 0; i < list.size(); i++) {
    if (searched == list.get(i)) {
        System.out.println("Value " + searched + " has index of: " + i);
    }
}
if (i == list.size()) {
    System.out.println("value " + searched + " is not found");
}

Carlos López Marí · Answer 5 · 2018-02-22T20:05:08.953

0

I would sugger doing this if you want to use .contains() method:

ArrayList<Integer> bufflist = list.clone(); // you create a temporal back of the list to operate with the same list without destroying the data
int searched = Integer.parseInt(reader.nextLine());

while (bufflist.contains(searched)) {
    System.out.println("Value " + searched + " has index of: " + (list.indexOf(searched));
     listbuff.remove(list.indexOf(searched)); //you remove from the list so it doesn´t find it again

}

edited Feb 22 '18 at 20:05

answered Feb 21 '18 at 11:12

Carlos López Marí

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3
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Several issues here. You're not actually creating a new list with `bufflist`, but merely referencing and changing the original. Also, inside the loop you're removing the element and then searching for it using the `indexOf` method. Probably should switch the two lines inside the loop. – Henrik Aasted Sørensen Feb 21 '18 at 14:22
1

@Henrik Fixed it! – Carlos López Marí Feb 22 '18 at 20:05

score -1 · Answer 6 · answered Feb 21 '18 at 11:10

Try below code

import java.util.ArrayList;
import java.util.Scanner;

public class KysytynLuvunIndeksi {

public static void main(String[] args) {
    Scanner reader = new Scanner(System.in);

    ArrayList<Integer> list = new ArrayList<>();
    while (true) {
        int read = Integer.parseInt(reader.nextLine());
        if (read == -1) {
            break;
        }

        list.add(read);
    }

    System.out.print("What are you looking for? ");
    int searched = Integer.parseInt(reader.nextLine());
    for (int i : list) {
        if (i == searched) {
            System.out.println("Value " + searched + " has index of: " + (list.indexOf(searched)));
        }
    }
}

}

No, it'll just print the index of the first occurrence however many times the number appears in the list. — Kevin Anderson, Feb 21 '18 at 11:16
Yeah, this is basically repeating the original problem of the question. — Henrik Aasted Sørensen, Feb 21 '18 at 11:35

Java index of items in arrayList

6 Answers6