CODEWITHSUNDEEP
javaandroidoop
Nitish
Nitish

Reputation: 323

How to get Object from ArrayList Without loop?

Suppose I have a class Employee :

class Employee {
         Employee(String name, int age)
         {
             this.name = name ;
             this.age = age;
         }
         String name ;
         int age;
     }

Now Create a List of this like :

ArrayList<Employee> aa = new ArrayList<Employee>();
 aa.add(new Employee("Nitish", 26));
 aa.add(new Employee("Sh", 2));
 aa.add(new Employee("S", 1));

Can i get Employee Object Where name value is "Nitish"? Without For loop

Upvotes: 2

Views: 9266

Answers (4)

Aaron
Aaron

Reputation: 3894

I guess your interviewer just doesn't want you to use for or while loops to find objects in an ArrayList, but you can actually find them "without loops".

First, you need to override equals and hashCode of Employee class:

@Override public boolean equals(Object obj) {
    // ...
}

@Override public int hashCode() {
    // ...
}

Now you can find your object with ArrayList.indexOf (uses equals to find match) by creating a dummy reference:

Employee target = new Employee("Nitish", 26);
int index = employees.indexOf(target);

It's kinda silly but I guess some interviewers want us to think outside-of-the-box. Even though under the hood it's using loops, but if my interviewer asks me the same question, instead of saying no you can't, I'd use this example because I want to try my best just not to use "loops" as asked, and explains how it works behind the scene. Then afterwards I'd briefly come up with other better solutions, and hope that works!

Upvotes: 1

madu_dev
madu_dev

Reputation: 87

ArrayList work on index based structure.to ADD,READ,GET value from array need to index of the specific index. instead store data on ArrayList you can use Map. Using key can be access value from Map. 

 Map<String, Employee> employeeMap = new HashMap<>();

    employeeMap.put("Nitish", new Employee("Nitish", 26));
    employeeMap.put("Sh", new Employee("Sh", 2));
    employeeMap.put("S", new Employee("S", 1));

    System.out.println(employeeMap.get("Nitish").getName());
    System.out.println(employeeMap.get("Nitish").getAge());

Upvotes: 0

Evgeniy Dorofeev
Evgeniy Dorofeev

Reputation: 136012

Try this 

Employee emp = aa.stream().filter(e->e.name.equals("Nitish")).findAny().orElse(null);

this version returns null if not found

Upvotes: 1

forpas
forpas

Reputation: 164089

You can use a filtered stream:

Object[] array = aa
        .stream()
        .parallel()
        .filter((emp) -> emp.name.equals("Nitish"))
        .toArray();

if (array.length > 0) {
    Employee employee = (Employee)array[0];
    String name = employee.name;
    int age = employee.age;
    System.out.println("Found Employee with Name = " + name + " and Age = " + age);
} else {
    System.out.println("Not Found");
}

Upvotes: 0

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