我想出的最简单的解决办法非常直接。

它的工作原理是在矩形中找到离圆最近的点，然后比较距离。

您可以通过一些操作来完成所有这些操作，甚至可以避免使用平方根函数。

public boolean intersects(float cx, float cy, float radius, float left, float top, float right, float bottom)
{
   float closestX = (cx < left ? left : (cx > right ? right : cx));
   float closestY = (cy < top ? top : (cy > bottom ? bottom : cy));
   float dx = closestX - cx;
   float dy = closestY - cy;

   return ( dx * dx + dy * dy ) <= radius * radius;
}

就是这样!上面的解决方案假设原点在世界的左上方，x轴指向下方。

如果你想要一个解决移动的圆形和矩形之间碰撞的解决方案，这要复杂得多，并且包含在我的另一个答案中。

下面是我的C代码，用于解决球体和非轴对齐的盒子之间的碰撞。它依赖于我自己的几个库例程，但它可能对某些人有用。我在游戏中使用了它，效果非常好。

float physicsProcessCollisionBetweenSelfAndActorRect(SPhysics *self, SPhysics *actor)
{
    float diff = 99999;

    SVector relative_position_of_circle = getDifference2DBetweenVectors(&self->worldPosition, &actor->worldPosition);
    rotateVector2DBy(&relative_position_of_circle, -actor->axis.angleZ); // This aligns the coord system so the rect becomes an AABB

    float x_clamped_within_rectangle = relative_position_of_circle.x;
    float y_clamped_within_rectangle = relative_position_of_circle.y;
    LIMIT(x_clamped_within_rectangle, actor->physicsRect.l, actor->physicsRect.r);
    LIMIT(y_clamped_within_rectangle, actor->physicsRect.b, actor->physicsRect.t);

    // Calculate the distance between the circle's center and this closest point
    float distance_to_nearest_edge_x = relative_position_of_circle.x - x_clamped_within_rectangle;
    float distance_to_nearest_edge_y = relative_position_of_circle.y - y_clamped_within_rectangle;

    // If the distance is less than the circle's radius, an intersection occurs
    float distance_sq_x = SQUARE(distance_to_nearest_edge_x);
    float distance_sq_y = SQUARE(distance_to_nearest_edge_y);
    float radius_sq = SQUARE(self->physicsRadius);
    if(distance_sq_x + distance_sq_y < radius_sq)   
    {
        float half_rect_w = (actor->physicsRect.r - actor->physicsRect.l) * 0.5f;
        float half_rect_h = (actor->physicsRect.t - actor->physicsRect.b) * 0.5f;

        CREATE_VECTOR(push_vector);         

        // If we're at one of the corners of this object, treat this as a circular/circular collision
        if(fabs(relative_position_of_circle.x) > half_rect_w && fabs(relative_position_of_circle.y) > half_rect_h)
        {
            SVector edges;
            if(relative_position_of_circle.x > 0) edges.x = half_rect_w; else edges.x = -half_rect_w;
            if(relative_position_of_circle.y > 0) edges.y = half_rect_h; else edges.y = -half_rect_h;   

            push_vector = relative_position_of_circle;
            moveVectorByInverseVector2D(&push_vector, &edges);

            // We now have the vector from the corner of the rect to the point.
            float delta_length = getVector2DMagnitude(&push_vector);
            float diff = self->physicsRadius - delta_length; // Find out how far away we are from our ideal distance

            // Normalise the vector
            push_vector.x /= delta_length;
            push_vector.y /= delta_length;
            scaleVector2DBy(&push_vector, diff); // Now multiply it by the difference
            push_vector.z = 0;
        }
        else // Nope - just bouncing against one of the edges
        {
            if(relative_position_of_circle.x > 0) // Ball is to the right
                push_vector.x = (half_rect_w + self->physicsRadius) - relative_position_of_circle.x;
            else
                push_vector.x = -((half_rect_w + self->physicsRadius) + relative_position_of_circle.x);

            if(relative_position_of_circle.y > 0) // Ball is above
                push_vector.y = (half_rect_h + self->physicsRadius) - relative_position_of_circle.y;
            else
                push_vector.y = -((half_rect_h + self->physicsRadius) + relative_position_of_circle.y);

            if(fabs(push_vector.x) < fabs(push_vector.y))
                push_vector.y = 0;
            else
                push_vector.x = 0;
        }

        diff = 0; // Cheat, since we don't do anything with the value anyway
        rotateVector2DBy(&push_vector, actor->axis.angleZ);
        SVector *from = &self->worldPosition;       
        moveVectorBy2D(from, push_vector.x, push_vector.y);
    }   
    return diff;
}

球面和矩形相交于IIF 圆心和矩形的一个顶点之间的距离小于球体的半径 或 圆心与矩形的一条边之间的距离小于球面的半径([点线距离]) 或 圆的中心在矩形的内部 一点上距离:

P1 = [x1,y1]
P2 = [x2,y2]
Distance = sqrt(abs(x1 - x2)+abs(y1-y2))

点线路距离:

L1 = [x1,y1],L2 = [x2,y2] (two points of your line, ie the vertex points)
P1 = [px,py] some point

Distance d =  abs( (x2-x1)(y1-py)-(x1-px)(y2-y1) ) / Distance(L1,L2)

矩形内圆中心: 采用分离轴的方法:如果存在一个投影到一条直线上，将矩形与点分开，它们就不相交

您将点投影在平行于矩形边的直线上，然后可以很容易地确定它们是否相交。如果它们不在所有4个投影上相交，它们(点和矩形)就不能相交。

你只需要内积(x= [x1,x2]，y = [y1,y2]，x *y = x1*y1 + x2*y2)

你的测试应该是这样的:

//rectangle edges: TL (top left), TR (top right), BL (bottom left), BR (bottom right)
//point to test: POI

seperated = false
for egde in { {TL,TR}, {BL,BR}, {TL,BL},{TR-BR} }:  // the edges
    D = edge[0] - edge[1]
    innerProd =  D * POI
    Interval_min = min(D*edge[0],D*edge[1])
    Interval_max = max(D*edge[0],D*edge[1])
    if not (  Interval_min ≤ innerProd ≤  Interval_max ) 
           seperated = true
           break  // end for loop 
    end if
end for
if (seperated is true)    
      return "no intersection"
else 
      return "intersection"
end if

它没有假设一个轴对齐的矩形，并且很容易扩展用于测试凸集之间的交集。

我想出的最简单的解决办法非常直接。

它的工作原理是在矩形中找到离圆最近的点，然后比较距离。

您可以通过一些操作来完成所有这些操作，甚至可以避免使用平方根函数。

public boolean intersects(float cx, float cy, float radius, float left, float top, float right, float bottom)
{
   float closestX = (cx < left ? left : (cx > right ? right : cx));
   float closestY = (cy < top ? top : (cy > bottom ? bottom : cy));
   float dx = closestX - cx;
   float dy = closestY - cy;

   return ( dx * dx + dy * dy ) <= radius * radius;
}

就是这样!上面的解决方案假设原点在世界的左上方，x轴指向下方。

如果你想要一个解决移动的圆形和矩形之间碰撞的解决方案，这要复杂得多，并且包含在我的另一个答案中。

我为处理形状创建了类 希望你喜欢

public class Geomethry {
  public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;

    float rectCenterX = rectangleX + rectHalfWidth;
    float rectCenterY = rectangleY + rectHalfHeight;

    float deltax = Math.abs(rectCenterX - circleX);
    float deltay = Math.abs(rectCenterY - circleY);

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
        if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
}

public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;
    float rectHalfWidth2 = rectangleWidth2/2.0f;
    float rectHalfHeight2 = rectangleHeight2/2.0f;

    float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
    float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle
        if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
  } 
}

我有一个方法可以避免昂贵的毕达哥拉斯，如果没有必要的话。当矩形和圆的包围框不相交时。

对非欧几里得也适用

class Circle {
 // create the bounding box of the circle only once
 BBox bbox;

 public boolean intersect(BBox b) {
    // test top intersect
    if (lat > b.maxLat) {
        if (lon < b.minLon)
            return normDist(b.maxLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.maxLat, b.maxLon) <= normedDist;
        return b.maxLat - bbox.minLat > 0;
    }

    // test bottom intersect
    if (lat < b.minLat) {
        if (lon < b.minLon)
            return normDist(b.minLat, b.minLon) <= normedDist;
        if (lon > b.maxLon)
            return normDist(b.minLat, b.maxLon) <= normedDist;
        return bbox.maxLat - b.minLat > 0;
    }

    // test middle intersect
    if (lon < b.minLon)
        return bbox.maxLon - b.minLon > 0;
    if (lon > b.maxLon)
        return b.maxLon - bbox.minLon > 0;
    return true;
  }
}

minLat、maxLat可替换为minY、maxY, minLon、maxLon也可替换为minX、maxX normDist方法比全距离计算快一点。例如，在欧几里得空间中没有平方根(或者没有很多其他的haversine): dat =(lat-circleY);dLon = (lon-circleX);赋范= dLat * dLat + dLon * dLon。当然，如果你使用normDist方法你需要创建一个normedDist = dist*dist;对于圆来说

查看我的GraphHopper项目的完整的BBox和Circle代码。