using Map to implement efficient searching process - java

Question

I am creating a phone book API in Java. Every name in my phone book is mapped to a set of phone numbers & address. I want to build a search function which can search an entry in my phone book using a search parameter as a name or as the phone number itself.

My approach has been to add all my entries to a:

Map <String, PhoneNumber> book = new HashMap <String,PhoneNumber>();
book.put("Name1",new PhoneNumber(new Integer(12345),new Integer(123456));
book.get("Name1");
// PhoneNumber is my class which can have different types of phone numbers

I want to search by both the name which is the key and also by the value. I cannot do that using a HashMap. Is there a way to do it better to implement an efficient searching process?

Don't represent phone numbers as integers. You cannot represent leading zeros or # with an integer. — Thorbjørn Ravn Andersen
– Thorbjørn Ravn Andersen, Commented Sep 26, 2010 at 8:19
Good catch @thorbjorn. Further, phone numbers aren't really numbers - it makes no sense to do math on a phone number. Therefore, it isn't a number even if it looks like one. So we what see here could be a generic 'lookup' class, and a 'phone book' subclass that applies a 'the key must be a numeric string' rule. — Tony Ennis
– Tony Ennis, Commented Sep 26, 2010 at 12:21
@Tony, the "key must be numeric string" does not allow for #'s. — Thorbjørn Ravn Andersen
– Thorbjørn Ravn Andersen, Commented Sep 26, 2010 at 13:49
By '#' do you mean 'the key must be comprised of characters between 0 and 9 inclusive'? If so, I think we're saying the same thing. — Tony Ennis
– Tony Ennis, Commented Sep 26, 2010 at 14:06

Tony Ennis · Accepted Answer · 2010-09-26 02:58:59Z

2

Don't use a Map for your phone book. It's too limited. Instead create a PhoneBook class. Within that class have 2 Maps, one for the by-name searching and one for the by-number searching. Create an add() routine that adds Info to both maps.

public class PhoneBook {
   private Map<String,Info> byNumber;
   private Map<String,Info> byName;
}

Where "Info" is a class that tells you everything about the person.

answered Sep 26, 2010 at 2:58

Tony Ennis

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1 Comment

Tony Ennis Over a year ago

The above assumes names are unique. That's unrealistic of course. If they aren't, you'd likely have: private Map<String,List<Info>> byName; so you get a list of all people with that name.

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