我有以下代碼:算法找到矩形
int width = 10;
int height = 7;
bool[,] array1 = new bool[width, height];
string values =
"1100000000" +
"1100000011" +
"0001100011" +
"0001100000" +
"0001110000" +
"0000000110" +
"0000000110";
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
array1[x, y] = (values[x + y * width] == '1');
}
}
即時尋找一個算法,將提取,我們有一個1
所以從這個數據中我們會得到矩形 範圍(0 ,0,2,2), (8,1,2,2), (3,2,3,3), (7,5,2,2) 矩形的順序並不重要!
但我不知道如何做到這一點任何人有任何指針?
閱讀生鏽韋伯的回答後,我想出了以下內容:
private static List<Rectangle> GetRectangles(bool[,] array)
{
List<Rectangle> rectangles = new List<Rectangle>();
for (int x = 0; x < array.GetLength(0); x++)
{
for (int y = 0; y < array.GetLength(1); y++)
{
if (array[x, y])
{
rectangles.Add(GetRectangle(array, new Point(x, y)));
}
}
}
return rectangles;
}
static Rectangle GetRectangle(bool[,] array, Point startLocation)
{
int maxX = int.MinValue;
int minX = int.MaxValue;
int maxY = int.MinValue;
int minY = int.MaxValue;
HashSet<Point> visitedLocations = new HashSet<Point>();
Stack<Point> pointsToGo = new Stack<Point>();
Point location;
pointsToGo.Push(startLocation);
while (pointsToGo.Count > 0)
{
location = pointsToGo.Pop();
if (!location.X.IsBetween(0, array.GetLength(0) - 1))
continue;
if (!location.Y.IsBetween(0, array.GetLength(1) - 1))
continue;
if (!array[location.X, location.Y])
continue;
if (visitedLocations.Contains(location))
continue;
visitedLocations.Add(location);
pointsToGo.Push(new Point(location.X + 1, location.Y));
pointsToGo.Push(new Point(location.X, location.Y + 1));
pointsToGo.Push(new Point(location.X - 1, location.Y));
pointsToGo.Push(new Point(location.X, location.Y - 1));
}
foreach (Point location2 in visitedLocations)
{
array[location2.X, location2.Y] = false;
if (location2.X > maxX)
maxX = location2.X;
if (location2.X < minX)
minX = location2.X;
if (location2.Y > maxY)
maxY = location2.Y;
if (location2.Y < minY)
minY = location2.Y;
}
return new Rectangle(minX, minY, maxX - minX + 1, maxY - minY + 1);
}
public static bool IsBetween<T>(this T item, T start, T end)
{
return Comparer<T>.Default.Compare(item, start) >= 0
&& Comparer<T>.Default.Compare(item, end) <= 0;
}
您的值看起來有點偏離。爲了得到3,2,3,3,我認爲你需要在5,2和5,3的1。我會看看我能爲算法提出什麼。 – itsme86
實際上不是你的第3個矩形(3,2,2,2)? – kenny
你嘗試了什麼?當你找到一個塊(= 0-> 1或1-> 0轉換)時,你只需檢查下一行的索引是否匹配。當他們匹配時什麼也不做,當不匹配的時候彈出左上角矩形的開始和右下角的最後一行... –