Geometry 两个矩形之间的距离

如何计算两个矩形之间的距离？相交的距离应返回0

这里有一个快速函数，用于计算由CGSize表示的两个CGRECT之间的距离：

CGSize CGSizeDistanceBetweenRects(CGRect rect1, CGRect rect2)
{
    if (CGRectIntersectsRect(rect1, rect2))
    {
        return CGSizeMake(0, 0);
    }

    CGRect mostLeft = rect1.origin.x < rect2.origin.x ? rect1 : rect2;
    CGRect mostRight = rect2.origin.x < rect1.origin.x ? rect1 : rect2;

    CGFloat xDifference = mostLeft.origin.x == mostRight.origin.x ? 0 : mostRight.origin.x - (mostLeft.origin.x + mostLeft.size.width);
    xDifference = MAX(0, xDifference);

    CGRect upper = rect1.origin.y < rect2.origin.y ? rect1 : rect2;
    CGRect lower = rect2.origin.y < rect1.origin.y ? rect1 : rect2;

    CGFloat yDifference = upper.origin.y == lower.origin.y ? 0 : lower.origin.y - (upper.origin.y + upper.size.height);
    yDifference = MAX(0, yDifference);

    return CGSizeMake(xDifference, yDifference);
}

CGSize cgsizedInstanceBetweenRects（CGRect rect1，CGRect rect2）
{
if（CGRectIntersectsRect（rect1，rect2））
{
返回CGSizeMake（0,0）；
}
CGRect mostLeft=rect1.origin.x

在一个稍微相关的注释中，以下是如何计算两个给定CGRect的中心之间的距离：

CGFloat CGRectGetDistanceBetweenCenters( CGRect rect1, CGRect rect2 )
{
   CGPoint center1 = CGPointMake( CGRectGetMidX( rect1 ), CGRectGetMidY( rect1 ) );
   CGPoint center2 = CGPointMake( CGRectGetMidX( rect2 ), CGRectGetMidY( rect2 ) );

   CGFloat horizontalDistance = ( center2.x - center1.x );
   CGFloat verticalDistance = ( center2.y - center1.y );

   CGFloat distance = sqrt( ( horizontalDistance * horizontalDistance ) + ( verticalDistance * verticalDistance ) );

   return distance;
}

---

向批准的答案添加swift版本：

extension CGRect {

    func distance(from rect: CGRect) -> CGSize {
        if intersects(rect) {
            return CGSize(width: 0, height: 0)
        }

        let mostLeft = origin.x < rect.origin.x ? self : rect
        let mostRight = rect.origin.x < self.origin.x ? self : rect

        var xDifference = mostLeft.origin.x == mostRight.origin.x ? 0 : mostRight.origin.x - (mostLeft.origin.x + mostLeft.size.width)
        xDifference = CGFloat(max(0, xDifference))

        let upper = self.origin.y < rect.origin.y ? self : rect
        let lower = rect.origin.y < self.origin.y ? self : rect

        var yDifference = upper.origin.y == lower.origin.y ? 0 : lower.origin.y - (upper.origin.y + upper.size.height)
        yDifference = CGFloat(max(0, yDifference))

        return CGSize(width: xDifference, height: yDifference)
    }
}

扩展CGRect{ func距离（从rect:CGRect）->CGSize{ 如果相交（矩形）{ 返回CGSize（宽度：0，高度：0） } 设mostLeft=origin.x谢谢你的回答。但是假设长方形的大小不一样，我应该如何考虑尺寸差异？如果我理解你所问的，我不是舒尔，因为IMO。