如何检查一个点是否位于其他2个点之间的一条线上

Question

我该怎么写这个函数？任何例子都赞赏

function isPointBetweenPoints(currPoint, point1, point2):Boolean {

    var currX = currPoint.x;
    var currY = currPoint.y;

    var p1X = point1.x;
    var p1y = point1.y;

    var p2X = point2.x;
    var p2y = point2.y;

    //here I'm stuck
}

Answer 1

假设point1和point2不同,首先检查点是否在线上.为此你只需要一个矢量point1 -> currPoint和的"交叉产品" point1 -> point2.

dxc = currPoint.x - point1.x;
dyc = currPoint.y - point1.y;

dxl = point2.x - point1.x;
dyl = point2.y - point1.y;

cross = dxc * dyl - dyc * dxl;

当且仅当cross等于零时,你的观点就在线上.

if (cross != 0)
  return false;

现在,正如你所知,这一点确实存在于线上,现在是时候检查它是否位于原始点之间.x如果线条比"垂直"更"水平",则可以通过比较坐标来轻松完成,否则进行y坐标

if (abs(dxl) >= abs(dyl))
  return dxl > 0 ? 
    point1.x <= currPoint.x && currPoint.x <= point2.x :
    point2.x <= currPoint.x && currPoint.x <= point1.x;
else
  return dyl > 0 ? 
    point1.y <= currPoint.y && currPoint.y <= point2.y :
    point2.y <= currPoint.y && currPoint.y <= point1.y;

注意,如果输入数据是完整的,则上述算法是完全积分的,即它不需要整数输入的浮点计算.计算时要小心潜在的溢出cross.

PS这个算法绝对精确,这意味着它会拒绝非常接近线但不精确在线上的点.有时这不是我们需要的.但这是一个不同的故事.

Answer 2

Distance(point1,currPoint)+Distance(currPoint,point2)==Distance(point1,point2)

但是如果你有浮点值,要小心,它们的情况有所不同......

Answer 3

这与 JavaScript 无关。尝试以下算法，其中点 p1=point1 和 p2=point2，第三个点是 p3=currPoint：

v1 = p2 - p1
v2 = p3 - p1
v3 = p3 - p2
if (dot(v2,v1)>0 and dot(v3,v1)<0) return between
else return not between

如果您想确保它也在 p1 和 p2 之间的线段上：

v1 = normalize(p2 - p1)
v2 = normalize(p3 - p1)
v3 = p3 - p2
if (fabs(dot(v2,v1)-1.0)<EPS and dot(v3,v1)<0) return between
else return not between