Java 两个矩形相互重叠吗？

我研究了这个问题，并按照答案用Java实现了我自己的版本。我觉得很接近。。。但仍然不正确。你能就这个错误给我一些建议吗

完整的源代码可以在以下位置找到：

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第一行在r1x2>=r2x2中有一个错误：

正确答案应该是：

boolean isInside = ((r1x1 >= r2x1) && (r1x2 <= r2x2) && (r1y1 >= r2y1) && (r1y2 <= r2y2));

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第一行在r1x2>=r2x2中有一个错误：

正确答案应该是：

boolean isInside = ((r1x1 >= r2x1) && (r1x2 <= r2x2) && (r1y1 >= r2y1) && (r1y2 <= r2y2));

我想应该是这样

boolean isOverLap = (r1x1 < r2x2) && (r1x2 > r2x1) &&  (r1y1 < r2y2) && (r1y2 > r2y1);

更容易阅读，没有所有的否定

您可以将其视为四种情况的相反情况，每种情况都保证不会发生重叠：

boolean isNonOverLap = (r1x1 >= r2x2) || (r1x2 <= r2x1) || (r1y1 >= r2y2) || (r1y2 <= r2y1);

我想应该是这样

boolean isOverLap = (r1x1 < r2x2) && (r1x2 > r2x1) &&  (r1y1 < r2y2) && (r1y2 > r2y1);

更容易阅读，没有所有的否定

您可以将其视为四种情况的相反情况，每种情况都保证不会发生重叠：

boolean isNonOverLap = (r1x1 >= r2x2) || (r1x2 <= r2x1) || (r1y1 >= r2y2) || (r1y2 <= r2y1);

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两个矩形将重叠。。。。当相应的对角线具有相同的长度时

例如：

R1:x1，y1，x2，y2，x3，y3，x4，y4

R2:a1、b1、a2、b2、a3、b3、a4、b4

所以这应该是真的：

距离X1，y1，x3，y3=距离A1，b1，a3，b3

而且

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距离X2，y2，x4，y4=距离A2，b2，a4，b4两个直肠将重叠。。。。当相应的对角线具有相同的长度时

例如：

R1:x1，y1，x2，y2，x3，y3，x4，y4

R2:a1、b1、a2、b2、a3、b3、a4、b4

所以这应该是真的：

距离X1，y1，x3，y3=距离A1，b1，a3，b3

而且

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距离x2，y2，x4，y4=距离a2，b2，a4，b4这是我的解决方案，我也测试了它

package small_Progs;

class Point {
    int x, y;

    Point(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

class Rectangle {
    Point lt, lb, rt, rb;

    Rectangle(Point lt, Point lb, Point rt, Point rb) {
        this.lt = lt;
        this.lb = lb;
        this.rt = rt;
        this.rb = rb;
    }
}

public class OverlappingRectagles {

    public static void main(String arg[]) {
        Point lt1 = new Point(3, 8);
        Point lb1 = new Point(3, 5);
        Point rt1 = new Point(6, 8);
        Point rb1 = new Point(6, 5);

        Point lt2 = new Point(5, 6);
        Point lb2 = new Point(5, 3);
        Point rt2 = new Point(9, 6);
        Point rb2 = new Point(9, 3);

        Point lt3 = new Point(3, 7);
        Point lb3 = new Point(3, 6);
        Point rt3 = new Point(5, 7);
        Point rb3 = new Point(5, 6);

        Point lt4 = new Point(1, 2);
        Point lb4 = new Point(1, 1);
        Point rt4 = new Point(2, 2);
        Point rb4 = new Point(2, 1);

        Rectangle r1 = new Rectangle(lt1, lb1, rt1, rb1);
        Rectangle r2 = new Rectangle(lt2, lb2, rt2, rb2);
        Rectangle r3 = new Rectangle(lt3, lb3, rt3, rb3);
        Rectangle r4 = new Rectangle(lt4, lb4, rt4, rb4);

        OverlappingRectagles obj = new OverlappingRectagles();
        obj.isOverLapping(r1, r2);
        obj.isOverLapping(r1, r3);
        obj.isOverLapping(r1, r4);

    }

    private void isOverLapping(Rectangle rect1, Rectangle rect2) {
        Point l1 = rect1.lt;
        Point l2 = rect2.lt;

        Point r1 = rect1.rb;
        Point r2 = rect2.rb;

        if (l1.y < l2.y || l2.y < r1.y) {
            System.out.println("Not Overlapping");
        } else if (l1.x > r2.x || l2.x > r1.x) {
            System.out.println("Not Overlapping");
        } else {
            if ((l1.y > r2.y && l2.y > r1.y) || (l2.y > r1.y && r2.y > r2.y)) {
                System.out.println("Overlapping");
            } else {
                System.out.println("Not Overlapping");
            }

        }

    }
}

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这是我的解决方案，我也测试了它

package small_Progs;

class Point {
    int x, y;

    Point(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

class Rectangle {
    Point lt, lb, rt, rb;

    Rectangle(Point lt, Point lb, Point rt, Point rb) {
        this.lt = lt;
        this.lb = lb;
        this.rt = rt;
        this.rb = rb;
    }
}

public class OverlappingRectagles {

    public static void main(String arg[]) {
        Point lt1 = new Point(3, 8);
        Point lb1 = new Point(3, 5);
        Point rt1 = new Point(6, 8);
        Point rb1 = new Point(6, 5);

        Point lt2 = new Point(5, 6);
        Point lb2 = new Point(5, 3);
        Point rt2 = new Point(9, 6);
        Point rb2 = new Point(9, 3);

        Point lt3 = new Point(3, 7);
        Point lb3 = new Point(3, 6);
        Point rt3 = new Point(5, 7);
        Point rb3 = new Point(5, 6);

        Point lt4 = new Point(1, 2);
        Point lb4 = new Point(1, 1);
        Point rt4 = new Point(2, 2);
        Point rb4 = new Point(2, 1);

        Rectangle r1 = new Rectangle(lt1, lb1, rt1, rb1);
        Rectangle r2 = new Rectangle(lt2, lb2, rt2, rb2);
        Rectangle r3 = new Rectangle(lt3, lb3, rt3, rb3);
        Rectangle r4 = new Rectangle(lt4, lb4, rt4, rb4);

        OverlappingRectagles obj = new OverlappingRectagles();
        obj.isOverLapping(r1, r2);
        obj.isOverLapping(r1, r3);
        obj.isOverLapping(r1, r4);

    }

    private void isOverLapping(Rectangle rect1, Rectangle rect2) {
        Point l1 = rect1.lt;
        Point l2 = rect2.lt;

        Point r1 = rect1.rb;
        Point r2 = rect2.rb;

        if (l1.y < l2.y || l2.y < r1.y) {
            System.out.println("Not Overlapping");
        } else if (l1.x > r2.x || l2.x > r1.x) {
            System.out.println("Not Overlapping");
        } else {
            if ((l1.y > r2.y && l2.y > r1.y) || (l2.y > r1.y && r2.y > r2.y)) {
                System.out.println("Overlapping");
            } else {
                System.out.println("Not Overlapping");
            }

        }

    }
}

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boolean的错误是NOTOVERLAP=！isOverLap？：这是个谜吗？什么错误？标题编辑：你是，还是不是，我的选区？布尔值isNotOverLap=有什么问题！isOverLap？：这是个谜吗？什么错误？标题编辑：你是，还是你不是，我的选区？谢谢你指出，但我要求的是isOverLap和isNotOverLap…谢谢你指出，但我要求的是isOverLap和isNotOverLap。。。