Libgdx 2d楼动画

Question

我正在构建Runner Game。  地面等级。

public class Ground extends Actor {

private Texture texture;
private ArrayList<Vector2> groundLocations;
private float speed;
private float cameraLeft, cameraRight;
private int textureWidth;


public Ground() {
    this.cameraRight = Const.GAME_WIDTH + Const.GAME_MARGIN;
    this.cameraLeft = 0 - Const.GAME_MARGIN;
    texture = new Texture(Gdx.files.internal("ui/ground.png"));
    groundLocations = new ArrayList<Vector2>();
    init();
}

public void setSpeed(float speed) {
    this.speed = speed;
}

private void init() {

    textureWidth = texture.getWidth();
    float currentPosition = cameraLeft;
    while (currentPosition < cameraRight) {

        Vector2 newLocation = new Vector2(currentPosition, 0);
        groundLocations.add(newLocation);
        currentPosition += textureWidth;
    }

}

public int getFloorHeight() {
    return texture.getHeight();
}


@Override
public void act(float delta) {
    super.act(delta);
    int size = groundLocations.size();

    for (int i = 0; i < size; i++) {
        Vector2 location = groundLocations.get(i);
        location.x -= delta * speed;
        if (location.x < cameraLeft) {

            location.x = findMax().x + textureWidth;
        }
    }

}

private Vector2 findMax() {
    return Collections.max(groundLocations, new Vector2Comparator());
}


@Override
public void draw(Batch batch, float parentAlpha) {
    super.draw(batch, parentAlpha);
    for (Vector2 location : groundLocations) {
        batch.draw(texture, location.x, location.y);
    }
}




public void dispose() {
    if (texture != null)
        texture.dispose();
}

}

• 地面纹理为128x128

• GAME_WIDTH = 1024f

• GAME_MARGIN = 250f

• 速度=变化。

当地面根据速度和FPS移动时。 （速度*三角洲） 问题是：地面纹理之间总是存在差距。经过一定的运动 函数findMax找到具有最大X的纹理 任何帮助都会得到赞赏。

更新：图片

Answer 1

WOW Simply Solution无法相信我没有看到它。

   for (int i = 0; i < size; i++) {
    Vector2 location = groundLocations.get(i);
    if (location.x < cameraLeft) {

        location.x = findMax().x + textureWidth;
    }          
    location.x -= delta * speed;

}