用递归算法在迷宫中寻找最短路径

Question

我做了一个小的递归算法，以找到迷宫的解决方案，格式如下

###S###
##___##
##_#_##
#__#_##
#E___##

其中，“#”表示墙，“#”表示开放空间（可自由移动）。”S'表示开始位置，E'表示结束位置

我的算法运行得很好，但我想知道如何修改它以获得最短路径

/**
 * findPath()
 * 
 * @param location - Point to search
 * @return true when maze solution is found, false otherwise
 */
private boolean findPath(Point location) {
    // We have reached the end point, and solved the maze
    if (location.equals(maze.getEndCoords())) {
        System.out.println("Found path length: " + pathLength);
        maze.setMazeArray(mazeArray);
        return true;
    }

    ArrayList<Point> possibleMoves = new ArrayList<Point>();
    // Move Right
    possibleMoves.add(new Point(location.x + 1, location.y));
    // Down Move
    possibleMoves.add(new Point(location.x, location.y - 1));
    // Move Left
    possibleMoves.add(new Point(location.x - 1, location.y));
    // Move Up
    possibleMoves.add(new Point(location.x, location.y + 1));

    for (Point potentialMove : possibleMoves) {
        if (spaceIsFree(potentialMove)) {
            // Move to the free space
            mazeArray[potentialMove.x][potentialMove.y] = currentPathChar;
            // Increment path characters as alphabet
            if (currentPathChar == 'z')
                currentPathChar = 'a';
            else
                currentPathChar++;
            // Increment path length
            pathLength++;

            // Find the next path to traverse
            if (findPath(potentialMove)) {
                return true;
            }

            // Backtrack, this route doesn't lead to the end
            mazeArray[potentialMove.x][potentialMove.y] = Maze.SPACE_CHAR;
            if (currentPathChar == 'a')
                currentPathChar = 'z';
            else
                currentPathChar--;
            // Decrease path length
            pathLength--;
        }
    }

    // Previous space needs to make another move
    // We will also return false if the maze cannot be solved.
    return false;
}

在第一个街区中，我找到了路径并将其打断。还设置了写有路径的char[][]数组，该数组随后作为结果打印出来

它工作得很好，但我想知道什么是修改它的最佳方式，使它在找到第一条成功路径后不会崩溃，而是继续运行，直到找到可能的最短路径

我尝试过这样做，修改findPath（）方法并添加一个shortestPath和hasFoundPath变量。第一个表示到目前为止找到的最短路径的长度，hasFoundPath变量表示是否找到任何路径

    // We have reached the end point, and solved the maze
    if (location.equals(maze.getEndCoords())) {
        System.out.println("Found path length: " + pathLength);
        // Is this path shorter than the previous?
        if (hasFoundPath && pathLength < shortestPathLength) {
            maze.setMazeArray(mazeArray);
            shortestPathLength = pathLength;
        } else if (!hasFoundPath) {
            hasFoundPath = true;
            maze.setMazeArray(mazeArray);
            shortestPathLength = pathLength;            
        }
        //return true;
    }

但我还没能让它将Mazarray设置为它可能找到的任何最短路径的正确值

任何指导都将不胜感激：）谢谢

spaceIsFree（）方法只需确保上/左/下/右坐标在移动到它们之前有效。因此，它确保字符是一个“u”或“E”，并且不超出范围

Answer 1

这是我想出的BFS搜索解决方案。 它将起点标记为“1”，然后将它可以到达的每个相邻点标记为“2”，将可以到达的每个相邻点标记为“3”，依此类推

然后从末尾开始，使用递减的“level”值向后移动，从而得到最短路径

private LinkedList<Point> findShortestPath(Point startLocation) {
    // This double array keeps track of the "level" of each node.
    // The level increments, starting at the startLocation to represent the path
    int[][] levelArray = new int[mazeArray.length][mazeArray[0].length];

    // Assign every free space as 0, every wall as -1
    for (int i=0; i < mazeArray.length; i++)
        for (int j=0; j< mazeArray[0].length; j++) {
            if (mazeArray[i][j] == Maze.SPACE_CHAR || mazeArray[i][j] == Maze.END_CHAR)
                levelArray[i][j] = 0;
            else
                levelArray[i][j] = -1;
        }

    // Keep track of the traversal in a queue
    LinkedList<Point> queue = new LinkedList<Point>();
    queue.add(startLocation);

    // Mark starting point as 1
    levelArray[startLocation.x][startLocation.y] = 1;

    // Mark every adjacent open node with a numerical level value
    while (!queue.isEmpty()) {
        Point point = queue.poll();
        // Reached the end
        if (point.equals(maze.getEndCoords()))
            break;

        int level = levelArray[point.x][point.y];
        ArrayList<Point> possibleMoves = new ArrayList<Point>();
        // Move Up
        possibleMoves.add(new Point(point.x, point.y + 1));
        // Move Left
        possibleMoves.add(new Point(point.x - 1, point.y));
        // Down Move
        possibleMoves.add(new Point(point.x, point.y - 1));
        // Move Right
        possibleMoves.add(new Point(point.x + 1, point.y));

        for (Point potentialMove: possibleMoves) {
            if (spaceIsValid(potentialMove)) {
                // Able to move here if it is labeled as 0
                if (levelArray[potentialMove.x][potentialMove.y] == 0) {
                    queue.add(potentialMove);
                    // Set this adjacent node as level + 1
                    levelArray[potentialMove.x][potentialMove.y] = level + 1;
                }
            }
        }
    }
    // Couldn't find solution
    if (levelArray[maze.getEndCoords().x][maze.getEndCoords().y] == 0)
        return null;

    LinkedList<Point> shortestPath = new LinkedList<Point>();
    Point pointToAdd = maze.getEndCoords();

    while (!pointToAdd.equals(startLocation)) {
        shortestPath.push(pointToAdd);
        int level = levelArray[pointToAdd.x][pointToAdd.y];
        ArrayList<Point> possibleMoves = new ArrayList<Point>();
        // Move Right
        possibleMoves.add(new Point(pointToAdd.x + 1, pointToAdd.y));
        // Down Move
        possibleMoves.add(new Point(pointToAdd.x, pointToAdd.y - 1));
        // Move Left
        possibleMoves.add(new Point(pointToAdd.x - 1, pointToAdd.y));
        // Move Up
        possibleMoves.add(new Point(pointToAdd.x, pointToAdd.y + 1));

        for (Point potentialMove: possibleMoves)  {
            if (spaceIsValid(potentialMove)) {
                // The shortest level will always be level - 1, from this current node.
                // Longer paths will have higher levels.
                if (levelArray[potentialMove.x][potentialMove.y] == level - 1) {
                    pointToAdd = potentialMove;
                    break;
                }
            }
        }
    }

    return shortestPath;
}

spaceIsValid（）只是确保空间不超出边界

Answer 2

您的代码似乎执行了depth-first search（DFS）。要找到最短路径，您需要切换到breadth-first search（BFS）。这不是通过向现有代码中添加几个变量就能做到的。这需要重写你的算法

将DFS转换为BFS的一种方法是摆脱递归，转而使用显式的堆栈来跟踪到目前为止访问过的节点。在搜索循环的每次迭代中，你（1）从堆栈中弹出一个节点；（2） 检查该节点是否为解决方案；（3）将每个子对象推到堆栈上。在伪代码中，这看起来像：

深度优先搜索

stack.push(startNode)

while not stack.isEmpty:
    node = stack.pop()

    if node is solution:
        return
    else:
        stack.pushAll(node.children)

如果随后将堆栈切换到队列，这将隐式成为BFS，BFS自然会找到最短路径

宽度优先的serarch

queue.add(startNode)

while not queue.isEmpty:
    node = queue.remove()

    if node is solution:
        return
    else:
        queue.addAll(node.children)

还有几点需要注意：

1. 上述算法适用于树：没有回路的迷宫。如果你的迷宫有回路，那么你需要确保你不会重访你已经看到的节点。在这种情况下，需要添加逻辑来跟踪所有已访问的节点，并避免再次将它们添加到堆栈/队列中

2. 如前所述，这些算法将找到目标节点，但它们不记得到达目标节点的路径。补充说，这对读者来说是一个练习