有两种方式。
1. 排除法,排除四种不可能重叠的情况就是了。
1 public static boolean IsOverlap( Rectangle rect1, Rectangle rect2 ){ 2 float x1 = rect1.x, y1 = rect1.y, w1 = rect1.width, h1 = rect1.height; 3 float x2 = rect2.x, y2 = rect2.y, w2 = rect2.width, h2 = rect2.height; 4 5 return !( x1+w1<=x2 || x2+w2<=x1 || y1+h1<=y2 || y2+h2<=y1 ); 6 }
2. 直接法,保证两个矩形x和y方向都有重叠。
1 public static boolean IsOverlap( Rectangle rect1, Rectangle rect2 ){ 2 float x1 = rect1.x, y1 = rect1.y, w1 = rect1.width, h1 = rect1.height; 3 float x2 = rect2.x, y2 = rect2.y, w2 = rect2.width, h2 = rect2.height; 4 5 return x1<x2+w2 && x2<x1+w1 && y1<y2+h2 && y2<y1+h1; 6 }
两种方式都是只需要4次比较即可。