将IndexOf与arrayList中的customObject一起使用

Question

I'm wanting to be able to return the position of an object using the indexOf method, but only want to pass the name of the contact to search for this, is there any way for this to be done?

I currently have this method:

private static ArrayList<Contacts> contactList = new ArrayList<Contacts>();

public class Contacts {
private String name;
private String number;


public Contacts(String name, String number) {
    this.name = name;
    this.number = number;
}

public String getName() {
    return name;
}

public String getNumber() {
    return number;
}

public void setName(String name) {
    this.name = name;
}

public void setNumber(String number) {
    this.number = number;
}



public int findItem(String name) {

    return contactList.indexOf(name);
}

Answer 1

Heres a function that will achieve this without going through the whole list, I think the complexity is less than O(n):

public int findItem(String name)
    {
        int max = contactList.size();

        //you might have to subtract this by one 
        //I'm not sure off the top
        int descCnt = max;


        for(int cnt = 0; cnt <= max/2; cnt++)
        {
            if(contactList.get(cnt).getName().equals(name)) return cnt;
            if(contactList.get(descCnt).getName().equals(name)) return descCnt;
            --descCnt;
        }

    }

Answer 2

If you are doing a lot of lookups of Contacts by name, you can put the instances into a Map<String, Contacts> . The specific type of Map depends upon your requirements; a HashMap might suffice.

Instead of contactList.add(contacts) , you can use:

contactMap.put(contacts.getName(), contacts);

and then look up an item in the map using:

contactMap.get(someName);

This will be faster to do the lookups than scanning through a list each time: each lookup will be O(1) for a HashMap , compared to O(n) for a list. However, it uses more memory.

---

Incidentally, your Contacts class looks like it represents a single contact, so it should be named as a singular: Contact .

Also, your find method is currently declared as an instance method:

public int findItem(String name) {

meaning you actually need an instance of Contacts to find another instance of Contacts . Instead, declare it static :

public static int findItem(String name) {

then you can invoke it without an instance:

Contacts found = Contacts.find("name");

Answer 3

What you ask is not in the contract of List#indexOf(Object) , so no, you should not try to make the list work that way.

Instead, you can write your own method that will accomplish what you want relatively easily. Simply iterate over your list and find the Contact that matches the specified name.

/**
 * Returns the List index of the Contact with the specified name. If no such
 * Contact is found, -1 will be returned.
 */
public int findItem(String name) {
    for (int i = 0; i < contactList.size(); i++) {
        Contact contact = contactList.get(i);
        if (null == contact) continue;
        if (java.lang.Objects.equals(name, contact.getName())) return i;
    }
    return -1;
}

Answer 4

Just to add guys, I've been able to do it this way:

public void searchItem(String name) {
    for(int i = 0; i < contactList.size(); i++) {
        if(name.equals(contactList.get(i).getName())) {
            System.out.println("Found " + name);
            break;
        }
        else {
            System.out.println("Could not find name!");
        }
    }
}

However, is this not fairly inefficient if I were to have a larger list? Is there a more efficient way of doing this?

Answer 5

If you are interested. A better way is to override equals() and hashcode() in your object. And use the indexOf the proper way.

Your equals could determine the equality based on the name, therefore removing all that extra and unnecessary code.