定义2D方形瓦片的1D数组Resize问题：元素位置错位修复

需求说明 #

我有一个定义瓦片网格的1D数组，每个瓦片包含8×8个元素（元素值为0或1），网格的宽高可自定义。需要调整该数组的大小，支持向右或底部添加/移除瓦片，且调整后必须保证原有瓦片的元素位置正确对齐。

示例 #

原始1×1瓦片数组（8×8元素） #

每个角落的值为1：

{
1,0,0,0,0,0,0,1,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
1,0,0,0,0,0,0,1
}

期望的2×2瓦片数组（16×16元素） #

扩展后第一个瓦片的角落位置保持正确：

{
1,0,0,0,0,0,0,1,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
1,0,0,0,0,0,0,1,   0,0,0,0,0,0,0,0,

0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
}

现有尝试的问题 #

我尝试过多种缩放方法，但元素位置始终无法正确对齐。以下是最新的尝试代码：

static int[] ResizeSquare(int[] originalSquare, int originalWidth, int originalHeight, int newWidth, int newHeight)
{
    int elementsPerTile = 8 * 8;
    int[] resizedSquare = new int[newWidth * newHeight * elementsPerTile];

    for (int row = 0; row < Math.Min(originalHeight, newHeight); row++)
    {
        for (int col = 0; col < Math.Min(originalWidth, newWidth); col++)
        {
            int originalTileIndex = row * originalWidth + col;
            int newTileIndex = row * newWidth + col;

            for (int i = 0; i < elementsPerTile; i++)
            {
                // Calculate the position within the original and resized squares
                int originalPosition = originalTileIndex * elementsPerTile + i;
                int newPosition = newTileIndex * elementsPerTile + i;

                // Copy the original value to the corresponding position in the resized square
                resizedSquare[newPosition] = originalSquare[originalPosition];
            }
        }
    }

    return resizedSquare;
}

这段代码的运行结果不符合预期，元素位置错位：

{
1,0,0,0,0,0,0,1,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   1,0,0,0,0,0,0,1,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,

0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,   0,0,0,0,0,0,0,0,
}

正确实现方案 #

问题出在原代码是将整个瓦片作为一个块直接复制，忽略了大网格的行结构。正确的做法是按大网格的逐行逐列映射元素，确保每个瓦片内的行和列都对应到新网格的正确位置。

代码实现 #

static int[] ResizeSquare(int[] originalSquare, int originalWidth, int originalHeight, int newWidth, int newHeight)
{
    const int tileSize = 8;
    int newTotalElements = newWidth * newHeight * tileSize * tileSize;
    int[] resizedSquare = new int[newTotalElements];

    // 遍历新网格的每一行
    for (int gridRow = 0; gridRow < newHeight * tileSize; gridRow++)
    {
        // 计算当前行属于原网格的哪个瓦片行，以及瓦片内的行索引
        int originalTileRow = gridRow / tileSize;
        int innerTileRow = gridRow % tileSize;

        // 遍历新网格的每一列
        for (int gridCol = 0; gridCol < newWidth * tileSize; gridCol++)
        {
            // 计算当前列属于原网格的哪个瓦片列，以及瓦片内的列索引
            int originalTileCol = gridCol / tileSize;
            int innerTileCol = gridCol % tileSize;

            // 计算当前元素在新数组中的位置
            int newIndex = gridRow * newWidth * tileSize + gridCol;

            // 仅当当前位置属于原网格范围内时，复制值，否则保持默认0
            if (originalTileRow < originalHeight && originalTileCol < originalWidth)
            {
                // 计算该元素在原数组中的位置
                int originalTileIndex = originalTileRow * originalWidth + originalTileCol;
                int originalInnerIndex = innerTileRow * tileSize + innerTileCol;
                int originalIndex = originalTileIndex * tileSize * tileSize + originalInnerIndex;

                resizedSquare[newIndex] = originalSquare[originalIndex];
            }
        }
    }

    return resizedSquare;
}

逻辑说明 #

1. 以大网格的行和列为遍历单位，而非瓦片整体，确保每个元素的位置映射准确。
2. 对每个元素，计算它所属的瓦片以及在瓦片内的具体位置，再对应到原数组的索引。
3. 超出原网格范围的新增瓦片区域，保持默认值0。

这样处理后，原有瓦片的元素会精确对应到新网格的正确位置，符合预期效果。

内容的提问来源于stack exchange，提问作者Antônio Mateus