Question

我有以下函数，它应该产生笛卡尔平面中的所有坐标，我可以通过n步从原点到达：

原点是'位置'，步数是'强度'，这是一个int 1-10。但是，我不断收到stackoverflow错误。每次我调用它，然后在ArrayList位置调用clear。想法？

更新的代码：

// Returns all positions reachable in 'strength' steps
    public ArrayList<Int2D> findEscapeSpace(Int2D location, Field f) {
        // Are we still within the given radius?
        if((Math.abs(location.getX() - this.location.getX()) + Math.abs(location.getY() - this.location.getY())) < strength) {
            System.out.println("Starting on " + location);
            // If this position is not contained already, and if it doesn't contain a wall
            if(!positions.contains(location) && f.wallField.getObjectsAtLocation(location) == null) {
                positions.add(location);
                System.out.println("added " + location);
            }

            // Getting neighboring positions
            ArrayList<Int2D> neigh = findNeighPos(location, f);

            for(Int2D pos : neigh) {
                System.out.println("looking into " + pos + " at depth " + (Math.abs(location.getX() - this.location.getX()) + Math.abs(location.getY() - this.location.getY())) + " and strength " + strength);

                if(!positions.contains(pos))
                    findEscapeSpace(pos, f);

            }

        }
        System.out.println(positions.size());
        return positions;

    }

旧代码

public ArrayList<Int2D> positions = new ArrayList<Int2D>();

    // Returns all positions reachable in 'strength' steps
    public ArrayList<Int2D> findEscapeSpace(Int2D location, Field f) {

        // Are we still within the given radius?
        if((Math.abs(location.getX() - this.location.getX()) + Math.abs(location.getY() - this.location.getY())) < strength) {
            // If this position is not contained already, and if it doesn't contain a wall
            if(!positions.contains(location) && f.wallField.getObjectsAtLocation(location) == null)
                positions.add(location);

            // Getting neighboring positions
            ArrayList<Int2D> neigh = findNeighPos(location, f);

            for(Int2D pos : neigh) {

                findEscapeSpace(pos, f);

            }

        }

        return positions;

    }

public ArrayList<Int2D> findNeighPos(Int2D currentP, Field f) {

        ArrayList neighPositions = new ArrayList<Int2D>();

        int cx = currentP.getX();
        int cy = currentP.getY();

        int maxY = f.HEIGHT-1;
        int maxX = f.WIDTH-1;

        // A few checks to make sure we're not going off tack (literally)

        if(cx > 0 && cy < maxY)
            neighPositions.add(new Int2D(cx-1, cy+1));

        if(cy < maxY)
            neighPositions.add(new Int2D(cx, cy+1));

        if(cx < maxX && cy < maxY)
            neighPositions.add(new Int2D(cx+1, cy+1));

        if(cx > 0)
            neighPositions.add(new Int2D(cx-1, cy));

        if(cx < maxX)
            neighPositions.add(new Int2D(cx+1, cy));

        if(cx > 0 && cy > 0)
            neighPositions.add(new Int2D(cx-1, cy-1));

        if(cy > 0)
            neighPositions.add(new Int2D(cx, cy-1));

        if(cx < maxX && cy > 0)
            neighPositions.add(new Int2D(cx+1, cy-1));

        return neighPositions;

    }

Answer 1

您的递归似乎没有终止条件。看起来您可能希望将strength作为参数传递给findEscapeSpace()，并且当该方法递归时，它传递的值比传递给它的值少一个。

除此之外，您的算法看起来效率很低，因为它很可能每次多次生成和测试许多可到达的单元格，而且，检查每个单元格是否已被找到会比较昂贵。但那是下一个要解决的问题。

递归函数的StackOverflow

1 个答案: