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

为了它的价值，我使用这个mixin来计算谷歌Maps API v3应用程序的缠绕顺序。

该代码利用了多边形区域的副作用:顺时针旋转顺序的顶点产生一个正的区域，而逆时针旋转顺序的相同顶点产生一个负的区域。该代码还使用了谷歌Maps几何库中的一种私有API。我觉得使用它很舒服——使用风险自负。

示例用法:

var myPolygon = new google.maps.Polygon({/*options*/});
var isCW = myPolygon.isPathClockwise();

完整的单元测试示例@ http://jsfiddle.net/stevejansen/bq2ec/

/** Mixin to extend the behavior of the Google Maps JS API Polygon type
 *  to determine if a polygon path has clockwise of counter-clockwise winding order.
 *  
 *  Tested against v3.14 of the GMaps API.
 *
 *  @author  stevejansen_github@icloud.com
 *
 *  @license http://opensource.org/licenses/MIT
 *
 *  @version 1.0
 *
 *  @mixin
 *  
 *  @param {(number|Array|google.maps.MVCArray)} [path] - an optional polygon path; defaults to the first path of the polygon
 *  @returns {boolean} true if the path is clockwise; false if the path is counter-clockwise
 */
(function() {
  var category = 'google.maps.Polygon.isPathClockwise';
     // check that the GMaps API was already loaded
  if (null == google || null == google.maps || null == google.maps.Polygon) {
    console.error(category, 'Google Maps API not found');
    return;
  }
  if (typeof(google.maps.geometry.spherical.computeArea) !== 'function') {
    console.error(category, 'Google Maps geometry library not found');
    return;
  }

  if (typeof(google.maps.geometry.spherical.computeSignedArea) !== 'function') {
    console.error(category, 'Google Maps geometry library private function computeSignedArea() is missing; this may break this mixin');
  }

  function isPathClockwise(path) {
    var self = this,
        isCounterClockwise;

    if (null === path)
      throw new Error('Path is optional, but cannot be null');

    // default to the first path
    if (arguments.length === 0)
        path = self.getPath();

    // support for passing an index number to a path
    if (typeof(path) === 'number')
        path = self.getPaths().getAt(path);

    if (!path instanceof Array && !path instanceof google.maps.MVCArray)
      throw new Error('Path must be an Array or MVCArray');

    // negative polygon areas have counter-clockwise paths
    isCounterClockwise = (google.maps.geometry.spherical.computeSignedArea(path) < 0);

    return (!isCounterClockwise);
  }

  if (typeof(google.maps.Polygon.prototype.isPathClockwise) !== 'function') {
    google.maps.Polygon.prototype.isPathClockwise = isPathClockwise;
  }
})();

Sean的答案的JavaScript实现:

function calcArea(poly) { if(!poly || poly.length < 3) return null; let end = poly.length - 1; let sum = poly[end][0]*poly[0][1] - poly[0][0]*poly[end][1]; for(let i=0; i<end; ++i) { const n=i+1; sum += poly[i][0]*poly[n][1] - poly[n][0]*poly[i][1]; } return sum; } function isClockwise(poly) { return calcArea(poly) > 0; } let poly = [[352,168],[305,208],[312,256],[366,287],[434,248],[416,186]]; console.log(isClockwise(poly)); let poly2 = [[618,186],[650,170],[701,179],[716,207],[708,247],[666,259],[637,246],[615,219]]; console.log(isClockwise(poly2));

我很确定这是对的。这似乎是有效的:-)

这些多边形看起来是这样的，如果你想知道的话:

我将提出另一个解决方案，因为它很简单，不需要大量的数学运算，它只是使用了基本的代数。计算多边形的带符号面积。如果是负的，点是顺时针的，如果是正的，点是逆时针的。(这与Beta的解决方案非常相似。)

计算带符号的面积: A = 1/2 * (x1*y2 - x2*y1 + x2*y3 - x3*y2 +…+ xn*y1 - x1*yn)

或者在伪代码中:

signedArea = 0
for each point in points:
    x1 = point[0]
    y1 = point[1]
    if point is last point
        x2 = firstPoint[0]
        y2 = firstPoint[1]
    else
        x2 = nextPoint[0]
        y2 = nextPoint[1]
    end if

    signedArea += (x1 * y2 - x2 * y1)
end for
return signedArea / 2

注意，如果你只是检查顺序，你不需要麻烦除以2。

来源:http://mathworld.wolfram.com/PolygonArea.html

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。

虽然这些答案是正确的，但它们在数学上的强度比必要的要大。假设地图坐标，其中最北的点是地图上的最高点。找到最北的点，如果两个点相等，它是最北的，然后是最东的(这是lhf在他的答案中使用的点)。在你的观点中，

点[0]= (5,0)

点[1]= (6,4)

点[2]= (4,5)

点[3]= (1,5)

点[4]= (1,0)

If we assume that P2 is the most north then east point either the previous or next point determine clockwise, CW, or CCW. Since the most north point is on the north face, if the P1 (previous) to P2 moves east the direction is CW. In this case, it moves west, so the direction is CCW as the accepted answer says. If the previous point has no horizontal movement, then the same system applies to the next point, P3. If P3 is west of P2, it is, then the movement is CCW. If the P2 to P3 movement is east, it's west in this case, the movement is CW. Assume that nte, P2 in your data, is the most north then east point and the prv is the previous point, P1 in your data, and nxt is the next point, P3 in your data, and [0] is horizontal or east/west where west is less than east, and [1] is vertical.

if (nte[0] >= prv[0] && nxt[0] >= nte[0]) return(CW);
if (nte[0] <= prv[0] && nxt[0] <= nte[0]) return(CCW);
// Okay, it's not easy-peasy, so now, do the math
if (nte[0] * nxt[1] - nte[1] * nxt[0] - prv[0] * (nxt[1] - crt[1]) + prv[1] * (nxt[0] - nte[0]) >= 0) return(CCW); // For quadrant 3 return(CW)
return(CW) // For quadrant 3 return (CCW)