Gfrktyl

I wanted to discuss Dijkstra's implementation that I generally find on Internet, and then compare it with my own implementation.

On Baeldung and on Vogella (Thanks to both and many others, that I haven't referred here, for providing the program). What these programs basically do is first find the node with the minimal distance and then calculate the minimum distance! In this approach, we are looping more and also increasing unnecessary complexities...

I have written my Dijkstra implementation, and basically that's a BFS with a little bit of modification to find the minimum distance! I have given my program below and the crux to find the minimum distance is if (!distances.containsKey(thisCity)) { and the else part of it along with Map<City, Integer> distances = new HashMap<>();. My program gives the expected output without any problem. To me it looks easier implementation and also easier to understand.

I understand the same program can be written in thousand different ways but I just wanted to get it reviewed and wants opinion of fellow programmers... Thanks for your time and help

public int findMinDistance(City src, City dest) {

    Set<City> visited = new HashSet<>();



    Queue<City> queue = new LinkedList<>();



    Map<City, Integer> distances = new HashMap<>();

    distances.put(src, Integer.valueOf(0));

    visited.add(src);

    queue.add(src);



    City currentCity;

    while (queue.size() != 0) {

        currentCity = queue.poll();

        Set<City> cities = currentCity.getCityDistancesMap().keySet();

        for (City thisCity : cities) {

            int thisDistance = distances.get(currentCity).intValue() +

                    currentCity.getCityDistance(thisCity);

            if (!distances.containsKey(thisCity)) {

                distances.put(thisCity, thisDistance);

            } else {

                int oldDistance = distances.get(thisCity).intValue();

                if (oldDistance > thisDistance) {

                    distances.remove(thisCity);

                    distances.put(thisCity, thisDistance);

                }

            }

            if (!visited.contains(thisCity)) {

                visited.add(thisCity);

                queue.add(thisCity);

            }

        }

    }

    return distances.get(dest);

}