I have found the solution but wanted to ensure my logic is the most efficient. I feel that there is a better way. I have the (x,y) coordinate of the bottom left corner, height and width of 2 rectangles, and i need to return a third rectangle that is their intersection. I do not want to post the code as i feel it is cheating.
I may be over thinking this and writing inefficient code. I already turned in a working program but would like to find the best way for my own knowledge. If someone could either agree or point me in the right direction, that would be great!
You can also use the Rectangle source code to compare with your own algorithm:
/**
* Computes the intersection of this <code>Rectangle</code> with the
* specified <code>Rectangle</code>. Returns a new <code>Rectangle</code>
* that represents the intersection of the two rectangles.
* If the two rectangles do not intersect, the result will be
* an empty rectangle.
*
* @param r the specified <code>Rectangle</code>
* @return the largest <code>Rectangle</code> contained in both the
* specified <code>Rectangle</code> and in
* this <code>Rectangle</code>; or if the rectangles
* do not intersect, an empty rectangle.
*/
public Rectangle intersection(Rectangle r) {
int tx1 = this.x;
int ty1 = this.y;
int rx1 = r.x;
int ry1 = r.y;
long tx2 = tx1; tx2 += this.width;
long ty2 = ty1; ty2 += this.height;
long rx2 = rx1; rx2 += r.width;
long ry2 = ry1; ry2 += r.height;
if (tx1 < rx1) tx1 = rx1;
if (ty1 < ry1) ty1 = ry1;
if (tx2 > rx2) tx2 = rx2;
if (ty2 > ry2) ty2 = ry2;
tx2 -= tx1;
ty2 -= ty1;
// tx2,ty2 will never overflow (they will never be
// larger than the smallest of the two source w,h)
// they might underflow, though...
if (tx2 < Integer.MIN_VALUE) tx2 = Integer.MIN_VALUE;
if (ty2 < Integer.MIN_VALUE) ty2 = Integer.MIN_VALUE;
return new Rectangle(tx1, ty1, (int) tx2, (int) ty2);
}
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