这里有一个快速的单行测试:

if (length(max(abs(center - rect_mid) - rect_halves, 0)) <= radius ) {
  // They intersect.
}

这是轴对齐的情况，其中rect_二分之一是一个正向量，从矩形的中间指向一个角。length()中的表达式是一个从矩形中心到最近点的增量向量。这适用于任何维度。

有效，一周前才发现，现在才开始测试。

double theta = Math.atan2(cir.getX()-sqr.getX()*1.0,
                          cir.getY()-sqr.getY()*1.0); //radians of the angle
double dBox; //distance from box to edge of box in direction of the circle

if((theta >  Math.PI/4 && theta <  3*Math.PI / 4) ||
   (theta < -Math.PI/4 && theta > -3*Math.PI / 4)) {
    dBox = sqr.getS() / (2*Math.sin(theta));
} else {
    dBox = sqr.getS() / (2*Math.cos(theta));
}
boolean touching = (Math.abs(dBox) >=
                    Math.sqrt(Math.pow(sqr.getX()-cir.getX(), 2) +
                              Math.pow(sqr.getY()-cir.getY(), 2)));

我的方法:

从OBB /矩形上/中的圆计算closest_point (最近点将位于边缘/角落或内部) 计算从closest_point到圆心的squared_distance (距离的平方避免了平方根) 返回squared_distance <=圆半径的平方

我在制作这款游戏时开发了这个算法:https://mshwf.github.io/mates/

如果圆与正方形接触，那么圆的中心线与正方形中心线之间的距离应该等于(直径+边)/2。 让我们有一个名为touching的变量来保存这个距离。问题是:我应该考虑哪条中心线:水平的还是垂直的? 考虑这个框架:

每条中心线给出了不同的距离，只有一条是没有碰撞的正确指示，但利用人类的直觉是理解自然算法如何工作的开始。

They are not touching, which means that the distance between the two centerlines should be greater than touching, which means that the natural algorithm picks the horizontal centerlines (the vertical centerlines says there's a collision!). By noticing multiple circles, you can tell: if the circle intersects with the vertical extension of the square, then we pick the vertical distance (between the horizontal centerlines), and if the circle intersects with the horizontal extension, we pick the horizontal distance:

另一个例子，圆4:它与正方形的水平延伸相交，那么我们考虑水平距离等于接触。

Ok, the tough part is demystified, now we know how the algorithm will work, but how we know with which extension the circle intersects? It's easy actually: we calculate the distance between the most right x and the most left x (of both the circle and the square), and the same for the y-axis, the one with greater value is the axis with the extension that intersects with the circle (if it's greater than diameter+side then the circle is outside the two square extensions, like circle #7). The code looks like:

right = Math.max(square.x+square.side, circle.x+circle.rad);
left = Math.min(square.x, circle.x-circle.rad);

bottom = Math.max(square.y+square.side, circle.y+circle.rad);
top = Math.min(square.y, circle.y-circle.rad);

if (right - left > down - top) {
 //compare with horizontal distance
}
else {
 //compare with vertical distance
}

/*These equations assume that the reference point of the square is at its top left corner, and the reference point of the circle is at its center*/

这个函数检测Circle和Rectangle之间的碰撞(交集)。他的回答类似于e.James的方法，但这个方法检测矩形的所有角(不仅仅是右角)的碰撞。

注意:

aRect.origin.x和aRect.origin.y是矩形左下角的坐标!

aCircle。x和圆。y为圆心坐标!

static inline BOOL RectIntersectsCircle(CGRect aRect, Circle aCircle) {

    float testX = aCircle.x;
    float testY = aCircle.y;

    if (testX < aRect.origin.x)
        testX = aRect.origin.x;
    if (testX > (aRect.origin.x + aRect.size.width))
        testX = (aRect.origin.x + aRect.size.width);
    if (testY < aRect.origin.y)
        testY = aRect.origin.y;
    if (testY > (aRect.origin.y + aRect.size.height))
        testY = (aRect.origin.y + aRect.size.height);

    return ((aCircle.x - testX) * (aCircle.x - testX) + (aCircle.y - testY) * (aCircle.y - testY)) < aCircle.radius * aCircle.radius;
}

假设你有矩形的四条边，检查从这些边到圆心的距离，如果小于半径，那么这些形状是相交的。

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect