C#俄罗斯方块，清除形成的矩形而不是行或线

Question

我正在开发一款具有俄罗斯方块概念的游戏，但在我的游戏中我想要清除的是玩家形成的矩形而不是行或线。例如，我实例化了如果网格上有 4 行和 5 列填充了俄罗斯方块块，这些块将被清除。

到目前为止，这是我删除一行的代码：

using UnityEngine;
using System.Collections;

public class Grid : MonoBehaviour {

    // The Grid itself
    public static int w = 10;
    public static int h = 20;
    public static Transform[,] grid = new Transform[w, h];



    void Start () {

    }


     public static Vector2 roundVec2(Vector2 v) {
        return new Vector2(Mathf.Round(v.x),
                           Mathf.Round(v.y));
    }



    // helps to find out if a certain coordinate is in between the borders or if it's outside of the borders:

    public static bool insideBorder(Vector2 pos) {
        return ((int)pos.x >= 0 &&
                (int)pos.x < w &&
                (int)pos.y >= 0);
    }



    //The next function deletes all Blocks in a certain row. 

    public static void deleteRow(int y) {
        for (int x = 0; x < w; ++x)
        {
            Destroy(grid[x, y].gameObject);
            grid[x, y] = null;

        }
        Score.currentScore += 1;

    }




    // Whenever a row was deleted, the above rows should fall towards the bottom by one unit

    public static void decreaseRow(int y) {
        for (int x = 0; x < w; ++x)
        {
            if (grid[x, y] != null)
            {
                // Move one towards bottom
                grid[x, y - 1] = grid[x, y];
                grid[x, y] = null;

                // Update Block position
                grid[x, y - 1].position += new Vector3(0, -1, 0);
            }
        }
    }


    public static void decreaseRowsAbove(int y)
    {
        for (int i = y; i < h; ++i)
            decreaseRow(i);
    }



    // function that finds out if a row is full of blocks

    public static bool isRowFull(int y)
    {
        for (int x = 0; x < w; ++x)
            if (grid[x, y] == null)
                return false;
        return true;
    }


    //a function that deletes all full rows and then always decreases the above row's y coordinate by one.

    public static void deleteFullRows()
    {
        for (int y = 0; y < h; ++y)
        {
            if (isRowFull(y))
            {
                deleteRow(y);
                decreaseRowsAbove(y + 1);
                --y;

            }
        }
    }       
}

那么，如何清除俄罗斯方块中形成的矩形？

通常的俄罗斯方块删除行或行。但我想要的是清除一个直角棱镜。

感谢您的回复

Answer 1

像这样：

public static void deleteGrid(int startX, int startY, int width = 5, int height = 4) 
{

    // to avoid Errors, Check if indexes are in bound for grid.

    // if condition code goes here

    // then...

    for (int i = startX; i < startX + width; i++)
    {
        for (int j = startY; j < startY + height; j++) 
        {
            Destroy(grid[i, j].gameObject);
            grid[i, j] = null;
        }
    }
    Score.currentScore += 1;
}

这样称呼它：

Grid.deleteGrid(2,3); // default width = 5, height = 4 will be used here.

或者

Grid.deleteGrid(2,3,3,3); // this will delete 3X3 grid at index (2,3)

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