Efficient compare among vector elements

Question

Hi guys,
I've a basic question, I've solved it with lots of nester for and I'm sure it's not the best way to do it, so I hope you can help me.
I've a vector of object User, so Vector<User> and user object has a property age.
What I want is simply:
1) iterating this vector and make comparing of age field for each user with each user (for example, 3 user, user1 - user2, user2 -user3,user3-user1).
2) User older for each compare must be entered in a new vector.
So starting from a vector of n element I should have a new vector with dimension equal to n*(n-1)/2.
Could someone help me to find most efficient code? (please show some code example)
Thanks,
Regards

Answer 1

new_to_java123 wrote:
What I want is simply:
1) iterating this vector and make comparing of age field for each user with each user (for example, 3 user, user1 - user2, user2 -user3,user3-user1).
2) User older for each compare must be entered in a new vector.
So starting from a vector of n element I should have a new vector with dimension equal to n*(n-1)/2.Well what you want is simply a very inefficient way to solve the problem. As you describe it you will use an algorithm of complexity O(n**2), while it can be done in O(n)!
What you should do:
1) Parse the vector once and determine the minAge and maxAge
2) Create an array[min-age, max-age] which contains a list of users for each index. As this is not directly supported in Java, use an ArrayList of size (maxAge - minAge ++1) and populate it with an empty List for each index 0 .. maxAge - minAge. Each entry with index i in that ArrayList is a List of persons whose age is (i ++ minAge).
3) Now iterate over the original Vector and append each user to the List of persons of his age (in code: ageList.get(user.getAge() - minAge()).add(user); )
Voilá. Only two scans of the original Vector and you have a structure that holds
1) List of the users of a given age (ageList.get(age - minAge()).
2) All users older than the given age can be found by iterating the lists with greater index as (age - minAge()).
And of course the memory requirements are much lower also!
It needs to store only maxAge - minAge + 1 plus the outer ageList. if you store null instead of an empty list if no user has a specific age this can be further reduced at the cost of havein to do more null checks.