Solved

Java Hashmap:

Posted on 2003-11-20
10
802 Views
Last Modified: 2010-03-31
Hi

I am trying to write a simple hashmap program, to understand its use. I managed to add some objects to the hashmap, and some edges using this code. Now I would like to write a removeObject method and understand this involves testing if there are any edges then removing the object x set value from all the keysets. Could anyone look over my removeObject method here, tell me where I am going wrong?

Thanks

   ((Set)theMap.get(x)).add(y);
   ((Set)theMap.get(y)).add(x);



    HashMap theMap = new HashMap();

    boolean removeObject(Object x){
        boolean boolAns = containsObject(x);

        if(boolAns == true){
            ((Set)theMap.get(x));

            boolean boolEdge = containsEdge(x, y)

            if(boolEdge == true){
                Set keys = theMap.keySet();
                Iterator keyIter = keys.iterator();

                while(keyIter.hasNext){

                    //remove set values
                }

                theMap.remove(x);
            }
            else{
                theMap.remove(x);
            }
        }

        return boolAns;
    }

0
Comment
Question by:AntoniRyszard656
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
10 Comments
 
LVL 35

Expert Comment

by:TimYates
ID: 9785815
think you need to post a little more code...
0
 
LVL 7

Expert Comment

by:grim_toaster
ID: 9785944
Going through the code, I'm a little confused.  You have a call to ((Set)theMap.get(x)); which doesn't actually do anything, but I'm presuming you wanted to store the resultant Set into a variable?

What you then have is a call to another method (code would help) that determines if the passed map, and some undefined variable y, contains an edge.  If it does, then you completely empty the HashMap (which you could have done with theMap.clear(), unless there is some specific tidy-up code or something?), and then you try to remove the object from the HashMap, but if it contained an edge, the HashMap would now be empty.
0
 

Author Comment

by:AntoniRyszard656
ID: 9786255
I have posted the main methods, the program adds objects to a hashmap, and allows the objects to linked by an edge. The problem is when a object is removed from the hashmap, the edges are broken, and I think I would need to check in remove Object if the edge exists and remove all the links. I think:

x/y   y/x  y/x then I need to remove all the x links before removing the object x.



    void addObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns != true){
            theMap.put(x, new HashSet());
        }
    }

    boolean removeObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns == true){
            ((Set)theMap.get(x));

            boolean boolEdge = containsObjectEdge(x, y)

            if(boolEdge == true){
                Set keys = theMap.keySet();
                Iterator keyIter = keys.iterator();

                while(keyIter.hasNext){

                    //remove set values
                }

                theMap.remove(x);
            }
            else{
                theMap.remove(x);
            }
        }

        return boolAns;
    }

    void addObjectEdge(Object x, Object y){
        boolean boolAns = theMap.containsKey(x);

        if(boolAns == false){
            addObject(x);
        }

        boolAns = theMap.containsKey(y);

        if(boolAns == false){
            addObject(y);
        }

       ((Set)theMap.get(x)).add(y);
       ((Set)theMap.get(y)).add(x);
    }

0
Get 15 Days FREE Full-Featured Trial

Benefit from a mission critical IT monitoring with Monitis Premium or get it FREE for your entry level monitoring needs.
-Over 200,000 users
-More than 300,000 websites monitored
-Used in 197 countries
-Recommended by 98% of users

 

Author Comment

by:AntoniRyszard656
ID: 9787147
Can anyone offer guidance with removeObject method?
0
 

Author Comment

by:AntoniRyszard656
ID: 9789521
I have posted the main methods, the program adds objects to a hashmap, and allows the objects to linked by an edge. The problem is when a object is removed from the hashmap, the edges are broken, and I think I would need to check in remove Object if the edge exists and remove all the links. I think:

x/y   y/x  y/x then I need to remove all the x links before removing the object x.



    void addObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns != true){
            theMap.put(x, new HashSet());
        }
    }

    boolean removeObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns == true){
            ((Set)theMap.get(x));

            boolean boolEdge = containsObjectEdge(x, y)

            if(boolEdge == true){
                Set keys = theMap.keySet();
                Iterator keyIter = keys.iterator();

                while(keyIter.hasNext){

                    //remove set values
                }

                theMap.remove(x);
            }
            else{
                theMap.remove(x);
            }
        }

        return boolAns;
    }

    void addObjectEdge(Object x, Object y){
        boolean boolAns = theMap.containsKey(x);

        if(boolAns == false){
            addObject(x);
        }

        boolAns = theMap.containsKey(y);

        if(boolAns == false){
            addObject(y);
        }

       ((Set)theMap.get(x)).add(y);
       ((Set)theMap.get(y)).add(x);
    }

0
 

Author Comment

by:AntoniRyszard656
ID: 9789522
I have posted the main methods, the program adds objects to a hashmap, and allows the objects to linked by an edge. The problem is when a object is removed from the hashmap, the edges are broken, and I think I would need to check in remove Object if the edge exists and remove all the links. I think:

x/y   y/x  y/x then I need to remove all the x links before removing the object x.



    void addObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns != true){
            theMap.put(x, new HashSet());
        }
    }

    boolean removeObject(Object x){
        boolean boolAns = containsNode(x);

        if(boolAns == true){
            ((Set)theMap.get(x));

            boolean boolEdge = containsObjectEdge(x, y)

            if(boolEdge == true){
                Set keys = theMap.keySet();
                Iterator keyIter = keys.iterator();

                while(keyIter.hasNext){

                    //remove set values
                }

                theMap.remove(x);
            }
            else{
                theMap.remove(x);
            }
        }

        return boolAns;
    }

    void addObjectEdge(Object x, Object y){
        boolean boolAns = theMap.containsKey(x);

        if(boolAns == false){
            addObject(x);
        }

        boolAns = theMap.containsKey(y);

        if(boolAns == false){
            addObject(y);
        }

       ((Set)theMap.get(x)).add(y);
       ((Set)theMap.get(y)).add(x);
    }

0
 

Author Comment

by:AntoniRyszard656
ID: 9809330
Hello all,

I posted this question on hashmaps and hashsets several days ago, but I have never had any further code, after I posted more of the code as suggested.

I would much appreicate any guidance.
0
 
LVL 15

Expert Comment

by:jimmack
ID: 9823709
Hi AntoniRyszard656,

Sorry, but I'm a bit busy at the moment.  I'll try to get back to this question this evening (or tomorrow).  I've got the reference to your other question now, so if I've answered the original question you asked here, please accept one of my answers.  I'll continue in your other question when I get time.

Thanx.
0
 
LVL 15

Accepted Solution

by:
jimmack earned 50 total points
ID: 9842469
Sorry all, I should have included a link to the other question in my last comment.

http://www.experts-exchange.com/Programming/Programming_Languages/Java/Q_20793435.html
0
 
LVL 15

Expert Comment

by:jimmack
ID: 10066313
;-)
0

Featured Post

Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Are you developing a Java application and want to create Excel Spreadsheets? You have come to the right place, this article will describe how you can create Excel Spreadsheets from a Java Application. For the purposes of this article, I will be u…
In this post we will learn how to make Android Gesture Tutorial and give different functionality whenever a user Touch or Scroll android screen.
Video by: Michael
Viewers learn about how to reduce the potential repetitiveness of coding in main by developing methods to perform specific tasks for their program. Additionally, objects are introduced for the purpose of learning how to call methods in Java. Define …
This tutorial will introduce the viewer to VisualVM for the Java platform application. This video explains an example program and covers the Overview, Monitor, and Heap Dump tabs.
Suggested Courses

617 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question