1、移动矩形
2、判断一个点是否在矩形内部
3、求两个矩形合并后的矩形，通过函数返回值返回代表合并后矩形的一个新建立的矩形

4、求两个矩形交集，通过函数返回值返回代表两个矩形交集的一个新建立的矩形对象

2个回答

``````class Untitled {
public static void main(String[] args) {
Rectangle rect1 = new Rectangle(10, 10, 20, 10);
System.out.println(rect1);
rect1.move(15, 30);
System.out.println(rect1);
Rectangle rect2 = new Rectangle(5, 10, 15, 10);
System.out.println(rect2);
Rectangle rect3 = rect1.combine(rect2);
System.out.println(rect3);
System.out.println(rect3.isInside(30, 10));
System.out.println(rect3.isInside(100, 100));
System.out.println(rect1.intersectsWith(rect2.move(20, 30)));
}
}

class Rectangle
{
public int top;
public int left;
public int width;
public int height;
public Rectangle(int t, int l, int w, int h)
{
this.top = t;
this.left = l;
this.width = w;
this.height = h;
}

public Rectangle move(int x, int y)
{
top = x;
left = y;
return this;
}

public Boolean isInside(int x, int y)
{
return x > left && x < (left + width) && y > top && y < (top + height);
}

public Rectangle combine(Rectangle rect)
{
int x = left > rect.left ? rect.left : left;
int y = top > rect.top ? rect.top : top;
int x2 = left + width > rect.left + rect.width ? left + width : rect.left + rect.width;
int y2 = top + height > rect.top + rect.height ? top + height : rect.top + rect.height;
return new Rectangle(x, y, x2 - x, y2 - y);
}

public Rectangle intersectsWith(Rectangle rect)
{
int x = left > rect.left ? left : rect.left;
int y = top > rect.top ? top : rect.top;
int x2 = left + width > rect.left + rect.width ? rect.left + rect.width : left + width;
int y2 = top + height > rect.top + rect.height ? rect.top + rect.height : top + height;
if (x2 - x <= 0 || y2 - y <= 0) return null;
return new Rectangle(x, y, x2 - x, y2 - y);
}

@Override
public String toString()
{
return "top = " + top +
" left = " + left +
" width = " + width +
" height = " + height + "\n";
}

}
``````

top = 10 left = 10 width = 20 height = 10
top = 15 left = 30 width = 20 height = 10
top = 5 left = 10 width = 15 height = 10
top = 10 left = 5 width = 40 height = 20
false
false
top = 30 left = 20 width = 15 height = 5

1、移动矩阵直接改变矩阵四个点的坐标
2、点在矩阵内等价 ：点的坐标（x，y）在矩阵x跨域，y跨域内
3、坐标合并
4、各跨域的交集就是新矩阵的各个坐标位置