在数组中查找正方形

Question

我有一个充满 1 和 0 的二维数组，例如

    0 0 0 0 0 0 0 0 0 0
    0 0 1 1 1 1 1 1 0 0
    0 0 1 0 0 0 0 1 0 0
    0 0 1 0 0 0 0 1 0 0
    0 0 1 0 0 0 0 1 0 0
    0 0 1 1 1 1 1 1 0 0
    0 0 0 0 0 0 0 0 0 0

你可以看到数组中有一个正方形。 我正在尝试制作一个基于正方形制作矩形或矩形列表的函数。 所以这个例子会返回一个像

这样的矩形

rect.x = 2
rect.y = 1
rect.width = 7
rect.height = 5

这是我现在拥有的代码，但它没有返回任何内容

Dim rects As New List(Of Rectangle)
    For imgWidth As Integer = 0 To bow.GetUpperBound(0)
        For imgHeight As Integer = 0 To bow.GetUpperBound(1)
            If bow(imgWidth, imgHeight) = 1 Then

                If bow(imgWidth + 1, imgHeight) = 1 And 
                   bow(imgWidth + 2, imgHeight) = 1 And 
                   bow(imgWidth, imgHeight + 1) = 1 And 
                   bow(imgWidth, imgHeight + 2) = 1 Then

                    Dim r As New Rectangle

                    With r
                        .X = imgWidth
                        .Y = imgHeight
                    End With

                    For rectWidth As Integer = imgWidth To bow.GetUpperBound(0)
                        If bow(rectWidth, imgHeight) = 0 Then
                            r.Width = bow(rectWidth - 1, imgHeight)
                        End If
                    Next

                    For rectHeight As Integer = imgHeight To bow.GetUpperBound(1)
                        If bow(imgWidth, rectHeight) = 0 Then
                            r.Height = bow(rectHeight - 1, imgHeight)
                        End If
                    Next

                    rects.Add(r)
                End If
            End If
        Next
    Next

此外，数组必须能够有多个正方形。

Answer 1

我会这样做：

def rectangles(grid):
  rows = len(grid)
  cols = len(grid[0])

  hor_ones = [[0]] * rows
  for r in range(rows):
    for c in range(cols):
      hor_ones[r].append(hor_ones[r][c] + grid[r][c])

  ver_ones = [[0]] * cols
  for c in range(cols):
    for r in range(rows):
      ver_ones[c].append(ver_ones[c][r] + grid[r][c])

  ret = []
  for r1 in range(rows):
    for c1 in range(cols):
      for r2 in range(r1+1, rows):
        for c2 in range(c1+1, cols):
          if all_ones(hor_ones[r1], c1, c2) and all_ones(hor_ones[r2], c1, c2) and all_ones(ver_ones[c1], r1, r2) and all_ones(ver_ones[c2], r1, r2):
            ret.append((r1, c2, r2, c2))
  return ret

def all_ones(ones, x, y):
  return ones[y+1] - ones[x] == y - x + 1

注意：

• hor_ones[r][c] 是数量 前 c 列中的第 r 行。
• ver_ones[c][r] 是数量 前 r 行中 c 列中的那些。

因此，r 行中以及c1 和c2（含）列之间的个数为：

hor_ones[r][c2+1] - hor_ones[r][c1]

编辑

这是我在Java中的解决方案，也许你在VB.NET中更容易理解和实现：

public static List<Rectangle> findRectangles(int[][] grid) {
    int rows = grid.length;
    int cols = grid[0].length;

    int[][] horOnes = new int[rows][cols+1];
    for (int r = 0; r < rows; r++)
        for (int c = 0; c < cols; c++)
            horOnes[r][c+1] = horOnes[r][c] + grid[r][c];

    int[][] verOnes = new int[cols][rows+1];
    for (int c = 0; c < cols; c++)
        for (int r = 0; r < rows; r++)
            verOnes[c][r+1] = verOnes[c][r] + grid[r][c];

    List<Rectangle> ret = new ArrayList<Rectangle>();
    for (int r1 = 0; r1 < rows; r1++)
        for (int c1 = 0; c1 < cols; c1++)
            for (int r2 = r1+1; r2 < rows; r2++)
                for (int c2 = c1+1; c2 < cols; c2++)
                    if (allOnes(horOnes[r1], c1, c2) && allOnes(horOnes[r2], c1, c2) && allOnes(verOnes[c1], r1, r2) && allOnes(verOnes[c2], r1, r2))
                        ret.add(new Rectangle(r1, c1, r2, c2));
    return ret;
}

private static boolean allOnes(int[] ones, int x, int y) {
    return ones[y+1] - ones[x] == y - x + 1;
}