递归搜索房间的Arraylist中的路径

Question

我正在尝试在带有房间的地牢中找到一条小路，房间通过门连接起来。从一个房间到另一个房间总有一种方法。有人可以帮我构建一条路吗？我已经研究了路径搜索算法，但是由于只有一条可能的路径，所以它们很复杂，因此没有必要为房间增加重量。

这是我正在使用的Room对象：

public Room(int id, Tile centerTile, ArrayList<Room> neighbours, ArrayList<Door> doors) 

  id = unique room id

  centerTile = center of the room.

  neighbours = list of neighbouring rooms

  doors = list of doors leading to the neighbouring rooms

客房被存放在被称为探索房间的Arraylist中。

这是我到目前为止所得到的，但它不起作用。

package pathing;

import room.Room;
import room.RoomUtils;

import java.util.ArrayList;
import java.util.Iterator;

public class Path {

public static ArrayList<Room> findPath(Room start, Room target) {

    ArrayList<Room> frontier = new ArrayList<>();
    ArrayList<Node> visited = new ArrayList<>();
    ArrayList<Room> path = new ArrayList<>();

    frontier.add(RoomUtils.getRoomById(start.getId()));
    visited.add(new Node(RoomUtils.getRoomById(start.getId()), null));

    if (start.equals(target)) {
        path.add(RoomUtils.getRoomById(start.getId()));
    } else {
        Iterator frontierIterator = frontier.listIterator();
        while (frontierIterator.hasNext()) {
            Room front = (Room) frontierIterator.next();
            front = RoomUtils.getRoomById(front.getId());
            if (front.getNeighbours().size() > 0) {
                for (Room room : front.getNeighbours()) {
                    if (room.equals(target)) {
                        frontier.clear();
                        visited.add(new Node(RoomUtils.getRoomById(room.getId()), RoomUtils.getRoomById(front.getId())));
                        frontierIterator = frontier.listIterator();
                    } else {
                        frontier.remove(RoomUtils.getRoomById(front.getId()));
                        visited.add(new Node(RoomUtils.getRoomById(room.getId()), RoomUtils.getRoomById(front.getId())));
                        frontierIterator = frontier.listIterator();
                    }
                }
            }
        }

        Node startNode = visited.get(0);
        Node backtrackNode = getFrom(visited, target);

        if (backtrackNode != null && backtrackNode.getCurrent() != null) {
            path.add(backtrackNode.getCurrent());
        }

        while (backtrackNode != null && backtrackNode.getFrom() != null && !backtrackNode.equals(startNode)) {
            path.add(backtrackNode.getFrom());
            backtrackNode = getFrom(visited, backtrackNode.getFrom());
        }
    }
    return path;
}

private static Node getFrom(ArrayList<Node> nodes, Room from) {
    for (Node node : nodes) {
        if (node.getCurrent().equals(from)) {
            return node;
        }
    }
    return null;
}

private static class Node {
    private final Room current;
    private final Room from;

    public Node(Room current, Room from) {
        this.current = current;
        this.from = from;
    }

    public Room getCurrent() {
        return current;
    }

    public Room getFrom() {
        return from;
    }
}

}

Answer 1

你的问题的标题包含'递归'，所以我寻找一个递归的解决方案。对于每个房间，你可以说：如果我的一个邻居房间通往目标房间，那么我也是路径的一部分。当你的邻居都没有通往目标时，你也不会成为路径的一部分。

为了使它递归，我创建了一个从当前房间开始的新方法。访问的房间的总体路径和列表被定义为类变量，因此在调用checkRoom之后它们仍然可用。

package pathing;

import java.util.ArrayList;
import java.util.List;

import room.Room;

public class Path {

    // list of all visited rooms to detect loops
    List<Room> visited = new ArrayList<>();
    // path from start to the current room
    List<Room> path = new ArrayList<>();
    Room target;
    boolean pathFound=false;

    public void checkRoom(Room current) {
        if(pathFound) {
            return;
        }
        if(visited.contains(current))
        {
            // we've been here before, we have detected a loop
            return;
        }
        path.add(current);
        visited.add(current);
        if (current.equals(target)) {
            pathFound=true;
        } else {
            // there are always neighbours (at least the room we come from
            for (Room room : current.getNeighbours()) {
                checkRoom(room);
                if(pathFound) {
                    return;
                }
            }
            // all neighbours processed without success
            path.remove(current);
        }
    }

    public static List<Room> findPath(Room start, Room target) {
        Path p=new Path();
        p.target=target;
        p.checkRoom(start);
        return p.path;
    }

    public static void main (String args[])
    {
        //      2 - 3 - 4
        //      |   |
        //      1 - 5
        //      |
        //      6 - 7 - 9
        //          |
        //          8
        Room room1=new Room(1,new ArrayList<Room>());
        Room room2=new Room(2,new ArrayList<Room>());
        Room room3=new Room(3,new ArrayList<Room>());
        Room room4=new Room(4,new ArrayList<Room>());
        Room room5=new Room(5,new ArrayList<Room>());
        Room room6=new Room(6,new ArrayList<Room>());
        Room room7=new Room(7,new ArrayList<Room>());
        Room room8=new Room(8,new ArrayList<Room>());
        Room room9=new Room(9,new ArrayList<Room>());
        room1.getNeighbours().add(room2);
        room1.getNeighbours().add(room5);
        room1.getNeighbours().add(room6);
        room2.getNeighbours().add(room1);
        room2.getNeighbours().add(room3);
        room3.getNeighbours().add(room2);
        room3.getNeighbours().add(room4);
        room4.getNeighbours().add(room3);
        room5.getNeighbours().add(room3);
        room5.getNeighbours().add(room1);
        room6.getNeighbours().add(room1);
        room6.getNeighbours().add(room7);
        room7.getNeighbours().add(room6);
        room7.getNeighbours().add(room9);
        room7.getNeighbours().add(room8);
        List<Room> path=Path.findPath(room1, room9);
        for(Room room:path)
            System.out.println("Room:"+room.getId());
        }
}

编辑：小优化：每个房间总是至少有一个邻居，即我们来自的房间，所以不需要检查是否有邻居。当迷宫只有一个房间但在这种情况下，开始和目标在之前已经测试过的情况下是一个例外。