Java method to find the rectangle that is the intersection of two rectangles using only left bottom point, width and height?

Question

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, heig

Answer 1

You can also use the Rectangle source code to compare with your own algorithm:

/**
 * Computes the intersection of this Rectangle with the
 * specified Rectangle. Returns a new Rectangle
 * 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 Rectangle
 * @return    the largest Rectangle contained in both the
 *            specified Rectangle and in
 *            this Rectangle; 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);
}