Step 5 of 5

Finding in the list

Module 13 found the largest number in a list: keep the largest one so far in a variable, and replace it whenever a larger one comes along. With objects, the variable holds a whole object. After the loop you have the largest value and also whose it is:

Main.java

import java.util.ArrayList;

public class Main {
    public static void main(String[] args) {
        ArrayList<Pet> pets = new ArrayList<>();
        pets.add(new Pet("Rex", 4));
        pets.add(new Pet("Misty", 2));
        pets.add(new Pet("Kiwi", 11));

        System.out.println("Oldest: " + findOldest(pets));
        printPet(pets, "Misty");
        printPet(pets, "Bo");
    }

    public static Pet findOldest(ArrayList<Pet> pets) {
        Pet oldest = pets.get(0);
        for (Pet pet : pets) {
            if (pet.getAge() > oldest.getAge()) {
                oldest = pet;
            }
        }
        return oldest;
    }

    public static void printPet(ArrayList<Pet> pets, String name) {
        for (Pet pet : pets) {
            if (pet.getName().equals(name)) {
                System.out.println("Found: " + pet);
                return;
            }
        }
        System.out.println("No pet called " + name);
    }
}

Pet.java

public class Pet {
    private String name;
    private int age;

    public Pet(String name, int age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return this.name;
    }

    public int getAge() {
        return this.age;
    }

    public String toString() {
        return this.name + ", " + this.age + " years old";
    }
}

Output

Oldest: Kiwi, 11 years old
Found: Misty, 2 years old
No pet called Bo

Pet oldest = pets.get(0); starts from the first pet, which is always a real candidate. The loop compares every pet's age with the oldest one so far, and when a pet is older, it becomes the new oldest.
findOldest returns a Pet: a method can return an object, like any other value. The caller gets the whole pet, and can print it or call its getters.
With >, the first of two pets of the same age stays the oldest; with >=, the later one would win.
get(0) needs at least one pet in the list. When the list might be empty, check pets.isEmpty() first.

Finding one by name

printPet goes through the list until a pet's name equals the name it's looking for (with ==, Strings aren't compared reliably). The return ends the method as soon as the pet is found. Only when the loop gets through the whole list without finding it does the last line run.

Finding in the list

Finding one by name

Your task

Challenge 2

Challenge 3