Do these two rectangles overlap horizontally? - Java 2D Graphics

Java examples for 2D Graphics:Rectangle

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Description

Do these two rectangles overlap horizontally?

Demo Code

// All rights reserved.
//package com.java2s;
import java.awt.*;

public class Main {
    /**/*from  ww w  . j  a  v a  2s .c om*/
     * Private utility method.  Do these two rectangles overlap horizontally? <p>
     * NOTE:  This method will still return true if the two rectangles are positioned
     * one above the other even though they don't completely overlap.
     */
    private static boolean rectanglesOverlapping_XAxis(Rectangle rec1,
            Rectangle rec2) {
        boolean bOverlapping = false;
        if ((rec1 == null) || (rec2 == null)) {
            return false;
        }
        int nLeft1 = (int) (rec1.getX());
        int nRight1 = (int) (rec1.getX()) + (int) (rec1.getWidth());
        int nLeft2 = (int) (rec2.getX());
        int nRight2 = (int) (rec2.getX()) + (int) (rec2.getWidth());
        // Find the intersections. If two lines coming from a rectangle corner
        // both have intersection points then there IS overlap.
        boolean bLeft1_Intersects = false;
        boolean bRight1_Intersects = false;
        boolean bLeft2_Intersects = false;
        boolean bRight2_Intersects = false;
        // Are there intersections based on rectangle #1?
        if ((nLeft1 > nLeft2) && (nLeft1 < nRight2)) {
            bLeft1_Intersects = true;
        }
        if ((nRight1 > nLeft2) && (nRight1 < nRight2)) {
            bRight1_Intersects = true;
        }
        // Are there intersections based on rectangle #2?
        if ((nLeft2 > nLeft1) && (nLeft2 < nRight1)) {
            bLeft2_Intersects = true;
        }
        if ((nRight2 > nLeft1) && (nRight2 < nRight1)) {
            bRight2_Intersects = true;
        }
        // Do we overlap or not?
        if (bLeft1_Intersects || bLeft2_Intersects || bRight1_Intersects
                || bRight2_Intersects) {
            bOverlapping = true;
        }
        return bOverlapping;
    }
}

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