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 <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);
}

Answer 2

My variation of determining intersection of two rectangles in a small utility function.

//returns true when intersection is found, false otherwise.
//when returning true, rectangle 'out' holds the intersection of r1 and r2.
private static boolean intersection2(Rectangle r1, Rectangle r2,
        Rectangle out) {
    float xmin = Math.max(r1.x, r2.x);
    float xmax1 = r1.x + r1.width;
    float xmax2 = r2.x + r2.width;
    float xmax = Math.min(xmax1, xmax2);
    if (xmax > xmin) {
        float ymin = Math.max(r1.y, r2.y);
        float ymax1 = r1.y + r1.height;
        float ymax2 = r2.y + r2.height;
        float ymax = Math.min(ymax1, ymax2);
        if (ymax > ymin) {
            out.x = xmin;
            out.y = ymin;
            out.width = xmax - xmin;
            out.height = ymax - ymin;
            return true;
        }
    }
    return false;
}

Answer 3

Why not use JDK API to do this for you?

Rectangle rect1 = new Rectangle(100, 100, 200, 240);
Rectangle rect2 = new Rectangle(120, 80, 80, 120);
Rectangle intersection = rect1.intersection(rect2);

To use java.awt.Rectangle class, the parameters of the constructor are: x, y, width, height, in which x, y are the top-left corner of the rectangle. You can easily convert the bottom-left point to top-left.

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I recommend the above, but if you really want to do it yourself, you can follow the steps below:

say (x1, y1), (x2, y2) are bottom-left and bottom-right corners of Rect1 respectively, (x3, y3), (x4, y4) are those of Rect2.

• find the larger one of x1, x3 and the smaller one of x2, x4, say xL, xR respectively
  • if xL >= xR, then return no intersection else
• find the larger one of y1, y3 and the smaller one of y2, y4, say yT, yB respectively
  • if yT >= yB, then return no intersection else
  • return (xL, yB, xR-xL, yB-yT).

A more Java-like pseudo code:

// Two rectangles, assume the class name is `Rect`
Rect r1 = new Rect(x1, y2, w1, h1);
Rect r2 = new Rect(x3, y4, w2, h2);

// get the coordinates of other points needed later:
int x2 = x1 + w1;
int x4 = x3 + w2;
int y1 = y2 - h1;
int y3 = y4 - h2;

// find intersection:
int xL = Math.max(x1, x3);
int xR = Math.min(x2, x4);
if (xR <= xL)
    return null;
else {
    int yT = Math.max(y1, y3);
    int yB = Math.min(y2, y4);
    if (yB <= yT)
        return null;
    else
        return new Rect(xL, yB, xR-xL, yB-yT);
}

As you see, if your rectangle was originally defined by two diagonal corners, it will be easier, you only need to do the // find intersection part.