问题C:如何检测两条线段是否相交?

我也搜索过同样的话题，但我对答案并不满意。所以我写了一篇文章，非常详细地解释了如何检查两条线段是否与大量图像相交。这是完整的(并经过测试的)java代码。

以下是这篇文章，截取了最重要的部分:

检查线段a是否与线段b相交的算法如下所示:

什么是边界框?下面是两个线段的边界框:

如果两个边界框都有交点，则移动线段a，使其中一点在(0|0)处。现在你有了一条经过a定义的原点的直线，现在以同样的方式移动线段b，检查线段b的新点是否在直线a的不同两侧。如果是这样，则反过来检查。如果也是这样，线段相交。如果不相交，它们就不相交。

问题A:两条线段在哪里相交?

你知道两条线段a和b相交。如果你不知道，用我在C题中给你的工具检查一下。

现在你可以通过一些情况，并得到解决与七年级数学(见代码和交互示例)。

问题B:你如何检测两条线是否相交?

假设点A = (x1, y1)点B = (x2, y2) C = (x_3, y_3) D = (x_4, y_4) 第一行由AB定义(A != B)，第二行由CD定义(C != D)。

function doLinesIntersect(AB, CD) {
    if (x1 == x2) {
        return !(x3 == x4 && x1 != x3);
    } else if (x3 == x4) {
        return true;
    } else {
        // Both lines are not parallel to the y-axis
        m1 = (y1-y2)/(x1-x2);
        m2 = (y3-y4)/(x3-x4);
        return m1 != m2;
    }
}

问题D:两条直线在哪里相交?

检查问题B，它们是否相交。

直线a和b由每条直线上的两个点定义。 你基本上可以用和问题A相同的逻辑。

上面有很多解决方案，但我认为下面的解决方案很简单，很容易理解。

矢量AB和矢量CD相交当且仅当

端点a和b在线段CD的两边。 端点c和d在线段AB的对边。

更具体地说，a和b在线段CD的对面当且仅当两个三元组中有一个是逆时针顺序的。

Intersect(a, b, c, d)
 if CCW(a, c, d) == CCW(b, c, d)
    return false;
 else if CCW(a, b, c) == CCW(a, b, d)
    return false;
 else
    return true;

这里的CCW代表逆时针，根据点的方向返回真/假。

来源:http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/x06-sweepline.pdf 第二页

我尝试了很多方法，然后我决定自己写。就是这样:

bool IsBetween (float x, float b1, float b2)
{
   return ( ((x >= (b1 - 0.1f)) && 
        (x <= (b2 + 0.1f))) || 
        ((x >= (b2 - 0.1f)) &&
        (x <= (b1 + 0.1f))));
}

bool IsSegmentsColliding(   POINTFLOAT lineA,
                POINTFLOAT lineB,
                POINTFLOAT line2A,
                POINTFLOAT line2B)
{
    float deltaX1 = lineB.x - lineA.x;
    float deltaX2 = line2B.x - line2A.x;
    float deltaY1 = lineB.y - lineA.y;
    float deltaY2 = line2B.y - line2A.y;

    if (abs(deltaX1) < 0.01f && 
        abs(deltaX2) < 0.01f) // Both are vertical lines
        return false;
    if (abs((deltaY1 / deltaX1) -
        (deltaY2 / deltaX2)) < 0.001f) // Two parallel line
        return false;

    float xCol = (  (   (deltaX1 * deltaX2) * 
                        (line2A.y - lineA.y)) - 
                    (line2A.x * deltaY2 * deltaX1) + 
                    (lineA.x * deltaY1 * deltaX2)) / 
                 ((deltaY1 * deltaX2) - (deltaY2 * deltaX1));
    float yCol = 0;
    if (deltaX1 < 0.01f) // L1 is a vertical line
        yCol = ((xCol * deltaY2) + 
                (line2A.y * deltaX2) - 
                (line2A.x * deltaY2)) / deltaX2;
    else // L1 is acceptable
        yCol = ((xCol * deltaY1) +
                (lineA.y * deltaX1) -
                (lineA.x * deltaY1)) / deltaX1;

    bool isCol =    IsBetween(xCol, lineA.x, lineB.x) &&
            IsBetween(yCol, lineA.y, lineB.y) &&
            IsBetween(xCol, line2A.x, line2B.x) &&
            IsBetween(yCol, line2A.y, line2B.y);
    return isCol;
}

根据这两个公式:(由直线方程和其他公式简化而来)

找到两条线段的正确交点是一项具有大量边缘情况的非简单任务。下面是一个用Java编写的、有效的、经过测试的解决方案。

本质上，在求两条线段的交点时，有三种情况会发生:

线段不相交 有一个唯一的交点 交点是另一段

注意:在代码中，我假设x1 = x2和y1 = y2的线段(x1, y1)， (x2, y2)是有效的线段。从数学上讲，线段由不同的点组成，但为了完整起见，我在这个实现中允许线段作为点。

代码是从我的github回购

/**
 * This snippet finds the intersection of two line segments.
 * The intersection may either be empty, a single point or the
 * intersection is a subsegment there's an overlap.
 */

import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;

import java.util.ArrayList;
import java.util.List;

public class LineSegmentLineSegmentIntersection {

  // Small epsilon used for double value comparison.
  private static final double EPS = 1e-5;

  // 2D Point class.
  public static class Pt {
    double x, y;
    public Pt(double x, double y) {
      this.x = x; 
      this.y = y;
    }
    public boolean equals(Pt pt) {
      return abs(x - pt.x) < EPS && abs(y - pt.y) < EPS;
    }
  }

  // Finds the orientation of point 'c' relative to the line segment (a, b)
  // Returns  0 if all three points are collinear.
  // Returns -1 if 'c' is clockwise to segment (a, b), i.e right of line formed by the segment.
  // Returns +1 if 'c' is counter clockwise to segment (a, b), i.e left of line
  // formed by the segment.
  public static int orientation(Pt a, Pt b, Pt c) {
    double value = (b.y - a.y) * (c.x - b.x) - 
                   (b.x - a.x) * (c.y - b.y);
    if (abs(value) < EPS) return 0;
    return (value > 0) ? -1 : +1;
  }

  // Tests whether point 'c' is on the line segment (a, b).
  // Ensure first that point c is collinear to segment (a, b) and
  // then check whether c is within the rectangle formed by (a, b)
  public static boolean pointOnLine(Pt a, Pt b, Pt c) {
    return orientation(a, b, c) == 0 && 
           min(a.x, b.x) <= c.x && c.x <= max(a.x, b.x) && 
           min(a.y, b.y) <= c.y && c.y <= max(a.y, b.y);
  }

  // Determines whether two segments intersect.
  public static boolean segmentsIntersect(Pt p1, Pt p2, Pt p3, Pt p4) {

    // Get the orientation of points p3 and p4 in relation
    // to the line segment (p1, p2)
    int o1 = orientation(p1, p2, p3);
    int o2 = orientation(p1, p2, p4);
    int o3 = orientation(p3, p4, p1);
    int o4 = orientation(p3, p4, p2);

    // If the points p1, p2 are on opposite sides of the infinite
    // line formed by (p3, p4) and conversly p3, p4 are on opposite
    // sides of the infinite line formed by (p1, p2) then there is
    // an intersection.
    if (o1 != o2 && o3 != o4) return true;

    // Collinear special cases (perhaps these if checks can be simplified?)
    if (o1 == 0 && pointOnLine(p1, p2, p3)) return true;
    if (o2 == 0 && pointOnLine(p1, p2, p4)) return true;
    if (o3 == 0 && pointOnLine(p3, p4, p1)) return true;
    if (o4 == 0 && pointOnLine(p3, p4, p2)) return true;

    return false;
  }

  public static List<Pt> getCommonEndpoints(Pt p1, Pt p2, Pt p3, Pt p4) {

    List<Pt> points = new ArrayList<>();

    if (p1.equals(p3)) {
      points.add(p1);
      if (p2.equals(p4)) points.add(p2);

    } else if (p1.equals(p4)) {
      points.add(p1);
      if (p2.equals(p3)) points.add(p2);

    } else if (p2.equals(p3)) {
      points.add(p2);
      if (p1.equals(p4)) points.add(p1);

    } else if (p2.equals(p4)) {
      points.add(p2);
      if (p1.equals(p3)) points.add(p1);
    }

    return points;
  }

  // Finds the intersection point(s) of two line segments. Unlike regular line 
  // segments, segments which are points (x1 = x2 and y1 = y2) are allowed.
  public static Pt[] lineSegmentLineSegmentIntersection(Pt p1, Pt p2, Pt p3, Pt p4) {

    // No intersection.
    if (!segmentsIntersect(p1, p2, p3, p4)) return new Pt[]{};

    // Both segments are a single point.
    if (p1.equals(p2) && p2.equals(p3) && p3.equals(p4))
      return new Pt[]{p1};

    List<Pt> endpoints = getCommonEndpoints(p1, p2, p3, p4);
    int n = endpoints.size();

    // One of the line segments is an intersecting single point.
    // NOTE: checking only n == 1 is insufficient to return early
    // because the solution might be a sub segment.
    boolean singleton = p1.equals(p2) || p3.equals(p4);
    if (n == 1 && singleton) return new Pt[]{endpoints.get(0)};

    // Segments are equal.
    if (n == 2) return new Pt[]{endpoints.get(0), endpoints.get(1)};

    boolean collinearSegments = (orientation(p1, p2, p3) == 0) && 
                                (orientation(p1, p2, p4) == 0);

    // The intersection will be a sub-segment of the two
    // segments since they overlap each other.
    if (collinearSegments) {

      // Segment #2 is enclosed in segment #1
      if (pointOnLine(p1, p2, p3) && pointOnLine(p1, p2, p4))
        return new Pt[]{p3, p4};

      // Segment #1 is enclosed in segment #2
      if (pointOnLine(p3, p4, p1) && pointOnLine(p3, p4, p2))
        return new Pt[]{p1, p2};

      // The subsegment is part of segment #1 and part of segment #2.
      // Find the middle points which correspond to this segment.
      Pt midPoint1 = pointOnLine(p1, p2, p3) ? p3 : p4;
      Pt midPoint2 = pointOnLine(p3, p4, p1) ? p1 : p2;

      // There is actually only one middle point!
      if (midPoint1.equals(midPoint2)) return new Pt[]{midPoint1};

      return new Pt[]{midPoint1, midPoint2};
    }

    /* Beyond this point there is a unique intersection point. */

    // Segment #1 is a vertical line.
    if (abs(p1.x - p2.x) < EPS) {
      double m = (p4.y - p3.y) / (p4.x - p3.x);
      double b = p3.y - m * p3.x;
      return new Pt[]{new Pt(p1.x, m * p1.x + b)};
    }

    // Segment #2 is a vertical line.
    if (abs(p3.x - p4.x) < EPS) {
      double m = (p2.y - p1.y) / (p2.x - p1.x);
      double b = p1.y - m * p1.x;
      return new Pt[]{new Pt(p3.x, m * p3.x + b)};
    }

    double m1 = (p2.y - p1.y) / (p2.x - p1.x);
    double m2 = (p4.y - p3.y) / (p4.x - p3.x);
    double b1 = p1.y - m1 * p1.x;
    double b2 = p3.y - m2 * p3.x;
    double x = (b2 - b1) / (m1 - m2);
    double y = (m1 * b2 - m2 * b1) / (m1 - m2);

    return new Pt[]{new Pt(x, y)};
  }

}

下面是一个简单的用法示例:

  public static void main(String[] args) {

    // Segment #1 is (p1, p2), segment #2 is (p3, p4)
    Pt p1, p2, p3, p4;

    p1 = new Pt(-2, 4); p2 = new Pt(3, 3);
    p3 = new Pt(0, 0);  p4 = new Pt(2, 4);
    Pt[] points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point = points[0];

    // Prints: (1.636, 3.273)
    System.out.printf("(%.3f, %.3f)\n", point.x, point.y);

    p1 = new Pt(-10, 0); p2 = new Pt(+10, 0);
    p3 = new Pt(-5, 0);  p4 = new Pt(+5, 0);
    points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point1 = points[0], point2 = points[1];

    // Prints: (-5.000, 0.000) (5.000, 0.000)
    System.out.printf("(%.3f, %.3f) (%.3f, %.3f)\n", point1.x, point1.y, point2.x, point2.y);
  }

基于@Gareth Rees的回答，Python版本:

import numpy as np

def np_perp( a ) :
    b = np.empty_like(a)
    b[0] = a[1]
    b[1] = -a[0]
    return b

def np_cross_product(a, b):
    return np.dot(a, np_perp(b))

def np_seg_intersect(a, b, considerCollinearOverlapAsIntersect = False):
    # https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect/565282#565282
    # http://www.codeproject.com/Tips/862988/Find-the-intersection-point-of-two-line-segments
    r = a[1] - a[0]
    s = b[1] - b[0]
    v = b[0] - a[0]
    num = np_cross_product(v, r)
    denom = np_cross_product(r, s)
    # If r x s = 0 and (q - p) x r = 0, then the two lines are collinear.
    if np.isclose(denom, 0) and np.isclose(num, 0):
        # 1. If either  0 <= (q - p) * r <= r * r or 0 <= (p - q) * s <= * s
        # then the two lines are overlapping,
        if(considerCollinearOverlapAsIntersect):
            vDotR = np.dot(v, r)
            aDotS = np.dot(-v, s)
            if (0 <= vDotR  and vDotR <= np.dot(r,r)) or (0 <= aDotS  and aDotS <= np.dot(s,s)):
                return True
        # 2. If neither 0 <= (q - p) * r = r * r nor 0 <= (p - q) * s <= s * s
        # then the two lines are collinear but disjoint.
        # No need to implement this expression, as it follows from the expression above.
        return None
    if np.isclose(denom, 0) and not np.isclose(num, 0):
        # Parallel and non intersecting
        return None
    u = num / denom
    t = np_cross_product(v, s) / denom
    if u >= 0 and u <= 1 and t >= 0 and t <= 1:
        res = b[0] + (s*u)
        return res
    # Otherwise, the two line segments are not parallel but do not intersect.
    return None

如果矩形的每条边都是一条线段，并且用户绘制的部分也是一条线段，那么您只需检查用户绘制的线段是否与四条边线段相交。这应该是一个相当简单的练习，给定每个段的起点和终点。