Question

我正在尝试回答CodeForces上的1A剧院广场问题。它一直告诉我：＆＃34;测试1＆＃34;运行时错误

这是我的代码：

Animation dwn,up, left, right;
BufferedImage north, south, east, west;
Sprites sprites = new Sprites();
KeyManager keyManager = new KeyManager();
Rectangle boundsPlayer;
//Controller c;


boolean movUp, movDwn, movRight, movLeft;
String direction = "";



public Player(int x, int y){
    super(x,y);
    sprites.create();
    dwn = new Animation(100, sprites.player_down);
    up = new Animation(100, sprites.player_up);
    left = new Animation(100, sprites.player_left);
    right = new Animation(100, sprites.player_right);
    north = sprites.playerNorth;
    south = sprites.playerSouth;
    east = sprites.playerEast;
    west = sprites.playerWest;

    boundsPlayer = new Rectangle(0,0, 32, 32);
}

public void setX(int x) {
    this.x = x;
}

public void setY(int y) {
    this.y = y;
}

public Rectangle getBounds(){
    return new Rectangle((int) x, (int) y, 32, 32);
}

/**
 * player tick method, used to move the player
 */
public void update(){

    if(Game.getKeyManager().up){
        y -= 3;
        direction = "north";


    }
    if(Game.getKeyManager().down){
        y += 3;
        direction = "south";

    }
    if(Game.getKeyManager().left){
        x -= 3;
        direction = "west";

    }

    if(Game.getKeyManager().right){
        x += 3;
        direction = "east";
        //System.out.println(x);
    }

}

public int getX() {
    return (int)x;
}

public int getY() {
    return (int)y;
}

public String getDirection(){
    return direction;
}

/**
 * method used to return players facing direction sprite
 * @return
 */
public BufferedImage getPlayerDirection(){
    if(direction == "north"){
        return north;
    }

    else if(direction == "south"){
        return south;
    }

    else if(direction == "east"){
        return east;
    }

    else{
        return west;
    }

}


public void render(Graphics g){
    g.drawImage(getCurrentAnimationFrame(), (int) x, (int) y, null);

    //g.setColor(Color.red);
    //g.fillRect((int) x + boundsPlayer.x, (int) y + boundsPlayer.y, boundsPlayer.width, boundsPlayer.height);
}


/**
 * method used to return the bufferedImage of current frame
 * @return
 */
public BufferedImage getCurrentAnimationFrame(){
    if(Game.getKeyManager().right == true){
        return right.getCurrentFrame();
    }

    else if(Game.getKeyManager().left == true){
        return left.getCurrentFrame();
    }

    else if(Game.getKeyManager().up == true){
        return up.getCurrentFrame();
    }

    else if(Game.getKeyManager().down == true){
        return dwn.getCurrentFrame();
    }

    else {
        return getPlayerDirection();
    }


}

我还是初学者，所以如果您有任何意见或建议，我们将非常感谢并感谢您抽出时间阅读我的问题。

Answer 1

您将获得超出时间限制，因为您使用的概念有2个循环，即O（n）的时间复杂度。但实际的解决方案将具有O（1）的时间复杂度，即它不需要任何循环。

提示： - 您将需要，只需几个基于其他的解决方案。

1A - 剧院广场代码部队

1 个答案: