在Dijkstra算法中返回路径

Question

So, I've been implementing Dijkstra's algorithm for pathfinding through a maze I generate (I know some of you might think something like A* would be better, but Dijkstra's perfectly fits my needs), and I've run into a small issue. I feel like I'm overlooking something, but when I return the "path" from the start node to the end node, it returns the entire search path the algorithm took (every node it visited), and not the path to the node.

//Uses djikstra's algorithm to find the shortest path between two nodes
//"vertices" is the global ArrayList in the class with all vertices in the graph.
@Override
public ArrayList<Vertex> pathfind(Vertex v1, Vertex v2) {
    ArrayList<Vertex> path = new ArrayList<Vertex>();
    ArrayList<Vertex> unvisited = new ArrayList<Vertex>();
    ArrayList<Vertex> visited = new ArrayList<Vertex>();

    if (!vertices.contains(v1) || !vertices.contains(v2)) {
        return path;
    }

    //initialize distances
    v1.setDistance(0);
    for(Vertex vert : vertices) {
        if(vert != v1) {
            vert.setDistance(Integer.MAX_VALUE);
        }
        unvisited.add(vert);
    }
    Vertex current = v1;

    //begin
    while (!unvisited.isEmpty()) {  
        //for all adjacent vertices that are unvisited
        for (Vertex v : current.adjacentVertices()) {
            if (!visited.contains(v)) {
                //if the distance of that vertex is greater than 
                //the added distance of the current node + the edge connecting the two
                int pathDist = current.getDistance() + findEdge(current, v).getElement(); 
                if (v.getDistance() > pathDist) {
                    //assign the new distance
                    v.setDistance(pathDist);
                }
            }
        }

        //remove the current node from the visited set and add it to visited
        visited.add(current);
        unvisited.remove(current);

        //add current node to the path
        path.add(current);

        //return if we found the destination
        if (current == v2)) {
            return path;
        }

        //else move to the lowest value node
        current = findSmallest(unvisited);

    }
    return path;

}

I know it must be something painfully obvious but I'm hitting my head against the wall, any help would be appreciated. Thanks!

Answer 1

You might want to calculate the distances first/ stabilize the distance to minimum value using the algo, then run over the graph to get the nodes for the path.

As current = findSmallest(unvisited) might get you node from a path which is not connected.

Answer 2

谢谢大家的帮助，我最终制作了一个HashMap，其中包含指向先前节点的链接，并将其遍历到起点。