不要认为坐标表示像素的位置。把它们想象成像素之间。这样，2x2矩形的面积应该是4，而不是9。

bool bOverlap = !((A.Left >= B.Right || B.Left >= A.Right)
               && (A.Bottom >= B.Top || B.Bottom >= A.Top));

if (RectA.Left < RectB.Right && RectA.Right > RectB.Left &&
     RectA.Top > RectB.Bottom && RectA.Bottom < RectB.Top ) 

或者用笛卡尔坐标

(X1是左坐标，X2是右坐标，从左到右递增，Y1是上坐标，Y2是下坐标，从下到上递增——如果这不是你的坐标系统(例如，大多数计算机的Y方向是相反的)，交换下面的比较)……

if (RectA.X1 < RectB.X2 && RectA.X2 > RectB.X1 &&
    RectA.Y1 > RectB.Y2 && RectA.Y2 < RectB.Y1) 

假设你有矩形A和矩形B。 反证法是证明。四个条件中的任何一个都保证不存在重叠:

Cond1。如果A的左边在B的右边的右边， -那么A完全在B的右边 Cond2。如果A的右边在B的左边的左边， -那么A完全在B的左边 Cond3。如果A的上边在B的下边之下， -那么A完全低于B Cond4。如果A的下边在B的上边上面， -那么A完全高于B

不重叠的条件是

NON-Overlap => Cond1 Or Cond2 Or Cond3 Or Cond4

因此，重叠的充分条件是相反的。

Overlap => NOT (Cond1 Or Cond2 Or Cond3 Or Cond4)

德摩根定律说 不是(A或B或C或D)和不是A不是B不是C不是D是一样的 所以利用德·摩根，我们有

Not Cond1 And Not Cond2 And Not Cond3 And Not Cond4

这相当于:

A的左边到B的右边的左边，[RectA。左< RectB。正确的), A的右边到B的左边的右边，[RectA。对，>，RectB。左), A的顶部高于B的底部。Top > RectB。底), A的底部在B的顶部以下。底部< RectB。前)

Note 1: It is fairly obvious this same principle can be extended to any number of dimensions. Note 2: It should also be fairly obvious to count overlaps of just one pixel, change the < and/or the > on that boundary to a <= or a >=. Note 3: This answer, when utilizing Cartesian coordinates (X, Y) is based on standard algebraic Cartesian coordinates (x increases left to right, and Y increases bottom to top). Obviously, where a computer system might mechanize screen coordinates differently, (e.g., increasing Y from top to bottom, or X From right to left), the syntax will need to be adjusted accordingly/

struct Rect
{
   Rect(int x1, int x2, int y1, int y2)
   : x1(x1), x2(x2), y1(y1), y2(y2)
   {
       assert(x1 < x2);
       assert(y1 < y2);
   }

   int x1, x2, y1, y2;
};

//some area of the r1 overlaps r2
bool overlap(const Rect &r1, const Rect &r2)
{
    return r1.x1 < r2.x2 && r2.x1 < r1.x2 &&
           r1.y1 < r2.y2 && r2.x1 < r1.y2;
}

//either the rectangles overlap or the edges touch
bool touch(const Rect &r1, const Rect &r2)
{
    return r1.x1 <= r2.x2 && r2.x1 <= r1.x2 &&
           r1.y1 <= r2.y2 && r2.x1 <= r1.y2;
}

bool Square::IsOverlappig(Square &other)
{
    bool result1 = other.x >= x && other.y >= y && other.x <= (x + width) && other.y <= (y + height); // other's top left falls within this area
    bool result2 = other.x >= x && other.y <= y && other.x <= (x + width) && (other.y + other.height) <= (y + height); // other's bottom left falls within this area
    bool result3 = other.x <= x && other.y >= y && (other.x + other.width) <= (x + width) && other.y <= (y + height); // other's top right falls within this area
    bool result4 = other.x <= x && other.y <= y && (other.x + other.width) >= x && (other.y + other.height) >= y; // other's bottom right falls within this area
    return result1 | result2 | result3 | result4;
}

设这两个矩形是矩形A和矩形b，设它们的中心为A1和B1 (A1和B1的坐标很容易求出来)，设高为Ha和Hb，宽为Wa和Wb，设dx为A1和B1之间的宽度(x)， dy为A1和B1之间的高度(y)。

现在我们可以说我们可以说A和B重叠，当

if(!(dx > Wa+Wb)||!(dy > Ha+Hb)) returns true

如果矩形重叠，那么重叠面积将大于零。现在让我们求出重叠区域:

如果它们重叠，那么overlap-rect的左边缘将是max(r1。X1, r2.x1)和右边将是min(r1。x2, r2.x2)。所以重叠的长度是min(r1)X2 r2 X2 - max(r1。x1, r2.x1)

所以面积是:

area = (max(r1.x1, r2.x1) - min(r1.x2, r2.x2)) * (max(r1.y1, r2.y1) - min(r1.y2, r2.y2))

如果area = 0，则它们不重叠。

很简单，不是吗?