写一个函数判断两个矩形是否相交,并尽量优化运行速度。
bool IsOverlapped(int x1,int y1,int w1,int h1,int x2,int y2,int w2,int h2)其中w 是矩形的宽度,h是矩形的高度
bool IsOverlapped(int x1,int y1,int w1,int h1,int x2,int y2,int w2,int h2)其中w 是矩形的宽度,h是矩形的高度
public class RectIntersected { /**
* @param args
*/
public static void main(String[] args) {
RectIntersected ri = new RectIntersected();
Rectangle r1 = ri.getRect(5, 5, 5, 5);
Rectangle r2 = ri.getRect(4, 5, 5, 5);
System.out.println(ri.intersected(r1, r2)); }
public boolean intersected(Rectangle r1,Rectangle r2) {
if(r1.intersects(r2)) {
return true;
}
return false;
}
public Rectangle getRect(int x, int y,int w,int h) {
return new Rectangle(x,y,w,h);
}}用AWT类的RECTANGLE类就可以
public class RectIntersected { /**
* @param args
*/
public static void main(String[] args) {
RectIntersected ri = new RectIntersected();
Rectangle r1 = new Rectangle(5,5,5,5);
Rectangle r2 = new Rectangle(4,5,5,5);
System.out.println(ri.intersected(r1, r2)); }
public boolean intersected(Rectangle r1,Rectangle r2) {
if(r1.intersects(r2)) {
return true;
}
return false;
}
}