递归函数上的StackOverflow

Question

我有以下功能，该功能应该产生我可以在n步中从原点到达的笛卡尔平面中的所有坐标：

原点是“位置”，步数是“强度”，它是整数1-10。 但是，我不断收到stackoverflow错误。 每次调用它时，我都会在ArrayList位置上调用clear。 有什么想法吗？

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。 看起来您可能想要将strength作为参数传递给findEscapeSpace() ，并且当该方法递归使它传递的值比传递给它的值小一个时。

除此之外，您的算法看起来效率很低，因为它可能会多次生成并测试许多可到达的单元，此外，检查每个单元是否已被发现的成本也相对较高。 但这是下一个要克服的问题。