获得矩形向量的边界矩形的最有效方法(Most efficient way to get bounding rectangle of vector of rectangles)

 我有一个cv::Rects （来自opencv框架）的std::vector ，叫做rects ，但是我想同样的逻辑将应用于任何矩形对象，所以我不打算在这个问题中标记opencv。 
 
 我试图找到包含矢量中所有矩形的最小矩形，这是我迄今为止的： 
 
int t = min_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.tl().y < b.tl().y;})->tl().y;
int l = min_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.tl().x < b.tl().x;})->tl().x;
int b = max_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.br().y < b.br().y;})->br().y;
int r = max_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.br().x < b.br().x;})->br().x;
cv::Rect smallest_encompassing_rect = cv::Rect(cv::Point(t, l),cv::Point(b, r)))
 
 如果不清楚，则.tl()函数将获取矩形的左上角点，而.br()会在右下角。 
 
 我知道这会起作用，但我想知道是否有更有效的方法来解决问题。 我正在处理图像处理，所以我的代码是时间敏感的。 
 
 任何帮助将不胜感激，谢谢 

I have an std::vector of cv::Rects (from the opencv framework) called rects, but I suppose the same logic would be applied to any rectangle object, so I'm not going to tag opencv in this question.
 
I am trying to find the smallest rectangle that encompasses all the rectangles in the vector, this is what I have so far:
 
int t = min_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.tl().y < b.tl().y;})->tl().y;
int l = min_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.tl().x < b.tl().x;})->tl().x;
int b = max_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.br().y < b.br().y;})->br().y;
int r = max_element(rects.begin(), rects.end(), [](cv::Rect a, cv::Rect b){ a.br().x < b.br().x;})->br().x;
cv::Rect smallest_encompassing_rect = cv::Rect(cv::Point(t, l),cv::Point(b, r)))
 
If it's not clear, the .tl() function gets the top left point of the rectangle, and .br() gets the bottom right.
 
I know this will work, but I am wondering if there is a more efficient way to solve the problem. I am working with image processing, so my code is time sensitive.
 
Any help would be appreciated, Thanks

原文：https://stackoverflow.com/questions/42685732