我如何看到两个矩形在JavaScript还是伪代码中相交？

问题描述

它们由[x0，y0，x1，y1]对表示，它们由[x0，y0，x1，y1]表示矩形的左上角和右下角。或者你的解决方案可能是[x0，y0，宽度，高度]，如果它更简单一些，我可以调整我的函数的参数输入。

我试着看看是否有从矩形A的两个角落包含在矩形B中，但是如果A大于B并且B部分包含在A中，则它将表示它不重叠。现在我可以尝试A和B，但这似乎是一种不好的方式去做事情。

我无法预先制作一个大网格，并通过矩形占据单元格，因为它不知道矩形是什么。我所能说的是，它们是无符号整数，最小值为0，最大值未知。

解决方案

检查矩形绝对不是相交的。如果没有这些情况是真的，那么矩形必须相交。即：

  public boolean rectanglesIntersect（
 float minAx，float minAy，float maxAx，float maxAy，
 float minBx，float minBy，float maxBx，float maxBy）{
 boolean aLeftOfB = maxAx< minBx; 
 boolean aRightOfB = minAx> maxBx; 
 boolean aAboveB = minAy> maxBy; 
 boolean aBelowB = maxAy< minBy; 
 
 return！（aLeftOfB || aRightOfB || aAoveb || aBelowB）; 
} 
  

这说明了这个概念，但是可以通过内联布尔以利用 ||

的短路行为

I have two rectangles which I must return in a function whether they intersect or not.

They are represented by [ x0,y0, x1,y1 ] pairs that represent the top-left and bottom-right corner of the rectangles. Alternatively your solution could be [ x0, y0, width, height ] if its somehow simpler, I can adapt my function's parameter input by it.

I tried to see if any of the two corners from rectangle A are included in rectangle B but if A is larger than B and B is partially included in A, it will say it doesn't overlap. Now I could try A and B but this seems to be a bad way to do things.

I can't premake a big grid and occupy cells by the rectangles because it is unknown what rectangles come as. All I can tell is that they are unsigned integers, min 0 and with unknown max.

解决方案

Check for the cases where the rectangles are definitely not intersecting. If none of these cases are true then the rectangles must intersect. i.e.:

public boolean rectanglesIntersect(
      float minAx, float minAy, float maxAx, float maxAy,
      float minBx, float minBy, float maxBx, float maxBy ) {
    boolean aLeftOfB = maxAx < minBx;
    boolean aRightOfB = minAx > maxBx;
    boolean aAboveB = minAy > maxBy;
    boolean aBelowB = maxAy < minBy;

    return !( aLeftOfB || aRightOfB || aAboveB || aBelowB );
}

This illustrates the concept, but could be made slightly faster by inlining the booleans so as to take advantage of the short-circuiting behaviour of ||

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