Java 递归方法

我正在做一个砖块破解游戏，不知为什么我被一个讨厌的bug困住了，我无法找到它。游戏目前的状态：

下面是我想做的：假设我点击上图中圈出的红色按钮。我希望红色的砖块消失，红色上面的砖块适当地占据它们的位置

迄今为止的守则：

private void moveBrick(BrickHolder brickHolder) {

    Point brickHolderLocation = brickHolder.getBrickHolderLocation();

    Brick containedBrick = getBrickByXAndY(brickHolderLocation.x, brickHolderLocation.y); // getting the Brick at that location

    if (containedBrick == null) {
        // If in any case there should be no brick at that position, just go on with the Brick above
        if (brickHolderLocation.y == 0) { // Should we be at the top row, there's no need to continue
            return;
        } else {
            BrickHolder nextBrickHolder = getPanelByXAndY(brickHolderLocation.x, brickHolderLocation.y - 1);
            moveBrick(nextBrickHolder);
        }
    }

    if (brickHolderLocation.y == 0) { // Should we be at the top row, there's no need to continue
        return;
    }

    // Removing the current Contained Brick
    brickHolder.remove(containedBrick);

    // Getting the Brick I want to move, normally hosted at the above Panel
    Brick theOneToBeMoved = getBrickByXAndY(brickHolderLocation.x, brickHolderLocation.y - 1);

    if (theOneToBeMoved == null) {
        // If in any case the Panel above doesn't contain a Brick, then continue with the Panel above.
        BrickHolder nextBrickHolder = getPanelByXAndY(brickHolderLocation.x, brickHolderLocation.y - 1);
        moveBrick(nextBrickHolder);
    }

    // Getting the Panel above the current one, so that we may move the Brick hosted there,
    // To the current Panel
    BrickHolder toHoldTheNewBrick = getPanelByXAndY(brickHolderLocation.x, brickHolderLocation.y - 1);

    brickHolder.add(theOneToBeMoved); // Moving the Brick at the current Panel
    toHoldTheNewBrick.remove(theOneToBeMoved); // Removing that same brick from the Panel above
    theOneToBeMoved.setBrickLocation(brickHolderLocation); // Setting the Brick's new location.

    // Since we have gotten so far, we assume that everything worked perfectly and that it's time to continue
    // with the Panel above
    BrickHolder theNextOne = getPanelByXAndY(brickHolder.getBrickHolderLocation().x, brickHolder.getBrickHolderLocation().y - 1);

    moveBrick(theNextOne);
}

从我所做的调试来看，我相信问题就在这里：

if (brickHolderLocation.y == 0) { // Should we be at the top row, there's no need to continue
            return;
        }

一些关注点：

• Brick-我创建的扩展JButton的类。没别的了。思考 它是一个带有背景的普通按钮
• 砌砖工-一级 创建以承载砖块。这个类扩展了JPanel。唯一的 加法是一个（点）位置变量，为方便起见而添加 操纵

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编辑：谢谢大家！您的评论和/或回答为我指明了继续的正确道路

我没有足够的观点来评论，所以这更像是一个建议，但是如果你在最上面一排，你会不会仍然想移除砖块呢？这与您在调试中发现的问题相同。

我没有足够的观点来评论，因此这更像是一个建议，但是如果您在最上面一行，您是否仍然希望移除砖块？这与您在调试中发现的问题相同。

可能希望将此移到…我不确定我是否理解您的代码逻辑。你做两次传球吗？一个是移除相同颜色的相邻砖块，第二个是重新分配砖块？其逻辑是移除该JPanel中包含的按钮，然后使用上面的按钮将其移动到上一个JPanel，继续这样做，直到它到达最上面一行。你应该分两次这样做——1）首先移除所有相邻的红砖，然后2）在移除所有红砖后重新定位砖。我认为这将为您简化一些事情。为了更好地帮助您，请尽快发布，并可能希望将此移到…我不确定我是否理解您的代码逻辑。你做两次传球吗？一个是移除相同颜色的相邻砖块，第二个是重新分配砖块？其逻辑是移除该JPanel中包含的按钮，然后使用上面的按钮将其移动到上一个JPanel，继续这样做，直到它到达最上面一行。你应该分两次这样做——1）首先移除所有相邻的红砖，然后2）在移除所有红砖后重新定位砖。我认为这将为你简化事情。为了更好地帮助，请尽快发布，因此你建议我在移动JPanel后检查它的位置？我认为你应该使用brickHolder.remove（containedBrick）；在if（brickHolder.y==0）{return；}之前，您建议我在移动JPanel后检查它的位置是否正确？我认为您应该使用brickHolder.remove（containedBrick）；在if（brickHolderLocation.y==0）{return；}之前