以下是基于上述答案的swift 3.0解决方案:

    for (i, point) in allPoints.enumerated() {
        let nextPoint = i == allPoints.count - 1 ? allPoints[0] : allPoints[i+1]
        signedArea += (point.x * nextPoint.y - nextPoint.x * point.y)
    }

    let clockwise  = signedArea < 0

对于那些不想“重新发明轮子”的人，我认为值得一提的是，这个检查是在一个名为Shapely (github)的漂亮的Python包中实现的(它基于GEOS C/ c++库):

Shapely is a BSD-licensed Python package for manipulation and analysis of planar geometric objects. It is using the widely deployed open-source geometry library GEOS (the engine of PostGIS, and a port of JTS). Shapely wraps GEOS geometries and operations to provide both a feature rich Geometry interface for singular (scalar) geometries and higher-performance NumPy ufuncs for operations using arrays of geometries. Shapely is not primarily focused on data serialization formats or coordinate systems, but can be readily integrated with packages that are.

来源:https://shapely.readthedocs.io/en/stable/

一个给出OP坐标的小例子:

import numpy as np
from shapely.geometry import Polygon

points = np.array([
    (5,0),
    (6,4),
    (4,5),
    (1,5),
    (1,0)
])

P = Polygon(points)

这是新构造的多边形:

import matplotlib.pyplot as plt

x,y = P.exterior.coords.xy
plt.plot(x,y)
plt.axis('equal')
plt.grid()
plt.show()

你可以直接使用LinearRing的is_ccw属性来检查多边形是CW还是CCW:

type(P.exterior)
>: shapely.geometry.polygon.LinearRing

P.exterior.is_ccw
>: True

如果颠倒:

points = np.flipud(points)
points
>: 
array([[1, 0],
       [1, 5],
       [4, 5],
       [6, 4],
       [5, 0]])


P1 = Polygon(points)

P1.exterior.is_ccw
>: True

进一步阅读的文档和参考资料:

shaely is_ccw (github): https://github.com/shapely/shapely/blob/eba985c6e0170ecdd90c83592fd0afa7ae793cb8/shapely/predicates.py#L72-L108 Libgeos (github): https://github.com/libgeos/geos GEOS API参考:https://libgeos.org/doxygen/classgeos_1_1algorithm_1_1Orientation.html#a5af93795969b80f97d7997195974d7c8 GEOS实现(github): https://github.com/libgeos/geos/blob/ab0ce6dafdf7f75ec6d234b6c65bb209037dda17/src/algorithm/Orientation.cpp#L43-L133

以下是基于上述答案的swift 3.0解决方案:

    for (i, point) in allPoints.enumerated() {
        let nextPoint = i == allPoints.count - 1 ? allPoints[0] : allPoints[i+1]
        signedArea += (point.x * nextPoint.y - nextPoint.x * point.y)
    }

    let clockwise  = signedArea < 0

一个计算上更简单的方法，如果你已经知道多边形内的一个点:

从原始多边形中选择任意线段，按此顺序选择点及其坐标。 加上一个已知的“内部”点，形成一个三角形。 根据以上三点计算CW或CCW。

Javascript实现的lhf的答案 (再次强调，这只适用于简单的多边形，即不适用于图8)

let polygon = [ {x:5,y:0}, {x:6,y:4}, {x:4,y:5}, {x:1,y:5}, {x:1,y:0} ] document.body.innerHTML += `Polygon ${polygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(polygon)}` let reversePolygon = [] polygon.forEach(point=>reversePolygon.unshift(point)) document.body.innerHTML += `<br/>Polygon ${reversePolygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(reversePolygon)}` function isPolygonClockwise (polygon) { // From http://www.faqs.org/faqs/graphics/algorithms-faq/ "How do I find the orientation of a simple polygon?" // THIS SOMETIMES FAILS if the polygon is a figure 8, or similar shape where it crosses over itself // Take the lowest point (break ties with the right-most). if (polygon.length < 3) { return true // A single point or two points can't be clockwise/counterclockwise } let previousPoint = polygon[0] let lowestPoint = polygon[1] let nextPoint = polygon[2] polygon.forEach((point, index)=>{ if (point.y > lowestPoint.y || (point.y === lowestPoint.y && point.x > lowestPoint.x)) { // larger y values are lower, in svgs // Break ties with furthest right previousPoint = polygon[(index-1) >= (0) ? (index-1) : (polygon.length-1)] lowestPoint = polygon[index] nextPoint = polygon[(index+1) <= (polygon.length-1) ? (index+1) : (0)] } }) // Check the angle between the previous point, that point, and the next point. // If the angle is less than PI radians, the polygon is clockwise let angle = findAngle(previousPoint, lowestPoint, nextPoint) return angle < Math.PI } function findAngle(A,B,C) { var AB = Math.atan2(B.y-A.y, B.x-A.x); var BC = Math.atan2(C.y-B.y, C.x-B.x); if (AB < 0) AB += Math.PI*2 if (BC < 0) BC += Math.PI*2 return BC-AB; }

从其中一个顶点开始，计算每条边对应的角度。

第一个和最后一个将是零(所以跳过它们);对于其余部分，角度的正弦值将由归一化与(点[n]-点[0])和(点[n-1]-点[0])的单位长度的叉乘给出。

如果这些值的和是正的，那么你的多边形是逆时针方向绘制的。