Reputation: 5605
I have an 2 dimensional array that is full of 1s and 0s such as
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
You can see there is a square in th array. I am trying to make a function that will make a rectangle or list of rectangles based on the square. So the example would return a rectangle like
rect.x = 2
rect.y = 1
rect.width = 7
rect.height = 5
This is the code i have now but it just doe not return anything
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
Also the array must be able to have more than one square.
Upvotes: 4
Views: 542
Reputation: 3260
This is how I would do it:
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
Note that:
hor_ones[r][c]
is the number of
ones in row r in the first c columns.ver_ones[c][r]
is the number of
ones in column c in the first r rows.Therefore, the number of ones in row r
and between columns c1
and c2
(inclusive) is:
hor_ones[r][c2+1] - hor_ones[r][c1]
EDIT
Here's my solution in Java, maybe it is easier for you to understand and implement in VB.NET:
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;
}
Upvotes: 3