GLSL版:

// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
    float aside = dot((p - a),(b - a));
    if(aside< 0.0) return length(p-a);
    float bside = dot((p - b),(a - b));
    if(bside< 0.0) return length(p-b);
    vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
    return length(p - pointOnLine);
}

嘿，我昨天才写的。它在Actionscript 3.0中，基本上是Javascript，尽管你可能没有相同的Point类。

//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.  
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
    var nearestPt:Point; //closest point on seqment to pt

    var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
    var bLenSq:Number = dot( b, b ); //Segment length squared

    if( keyDot <= 0 )  //pt is "behind" st, use st
    {
        nearestPt = st  
    }
    else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
    {
        nearestPt = st.add(b);
    }
    else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
    {
        var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
        var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
        nearestPt = st.add(partOfB);
    }

    var dist:Number = (pt.subtract(nearestPt)).length;

    return dist;
}

此外，这里有一个关于这个问题的相当完整和可读的讨论:notejot.com

这里没有看到Java实现，所以我将Javascript函数从接受的答案转换为Java代码:

static double sqr(double x) {
    return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
    return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
    double l2 = dist2(v, w);
    if (l2 == 0) return dist2(p, v);
    double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0) return dist2(p, v);
    if (t > 1) return dist2(p, w);
    return dist2(p, new DoublePoint(
            v.x + t * (w.x - v.x),
            v.y + t * (w.y - v.y)
    ));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
    return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
    public double x;
    public double y;

    public DoublePoint(double x, double y) {
        this.x = x;
        this.y = y;
    }
}

一个2D和3D的解决方案

考虑基底的变化，使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上，最短的距离由

    u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u     -> d(B, C)

(到其中一个端点或到支撑线的距离，取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)

式中，d为AB线段的长度，u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。

AB.AC ≤ 0             -> |AC|
    0 ≤ AB.AC ≤ AB²   -> |ABxAC|/|AB|
          AB² ≤ AB.AC -> |BC|

这是我最后写的代码。这段代码假设一个点以{x:5, y:7}的形式定义。注意，这不是绝对最有效的方法，但它是我能想到的最简单、最容易理解的代码。

// a, b, and c in the code below are all points

function distance(a, b)
{
    var dx = a.x - b.x;
    var dy = a.y - b.y;
    return Math.sqrt(dx*dx + dy*dy);
}

function Segment(a, b)
{
    var ab = {
        x: b.x - a.x,
        y: b.y - a.y
    };
    var length = distance(a, b);

    function cross(c) {
        return ab.x * (c.y-a.y) - ab.y * (c.x-a.x);
    };

    this.distanceFrom = function(c) {
        return Math.min(distance(a,c),
                        distance(b,c),
                        Math.abs(cross(c) / length));
    };
}

请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。

几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。