判断点是否在凸多边形内

1,原理

假设凸多边形顶点,按照顺时针顺序构成顶点数组verts:Point[],依次取两个顶点构成线段序列。
若点落在凸多边形内,则必有:该点在所有的线段序列的右侧或者左侧。

2,补充几个概念

右手坐标系

让右手拇指指向x轴的正方向,食指指向y轴的正方向,如果中指能指向z轴的正方向,则称这个坐标系为右手直角坐标系。反之则是左手直角坐标系。
判断点是否在凸多边形内
另一种通俗的理解:
伸出右手,让拇指和食指成“L”形,大拇指向右,食指向上,其余的手指指向前方,这样就建立了一个右手坐标系。其中,拇指、食指和其余手指分别代表x,y,z轴的正方向。

向量点积

判断点是否在凸多边形内

判断点是否在凸多边形内

向量叉积

判断点是否在凸多边形内

二维下叉积公式

判断点是否在凸多边形内

3,源码

    // 是否顺时针
    private static isClockwise(a: Point, b: Point, c: Point): boolean {
        //取ab向量,ac向量,进行叉乘
        return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y) < 0;
    }
    private static isClockwiseMargin(a: Point, b: Point, c: Point): boolean {
        return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y) <= 0;
    }
    //verts,六边形的顶点,顺时针存储
    private static inConvexPolygon(p: Point, verts: Point[]): boolean {
        if (verts.length < 3) {
            return false;
        }

        let iniResult = this.isClockwise(verts[0], verts[1], p);
        for (let i = 1; i < verts.length; i++) {
            let p1 = verts[i];
            let p2 = null;
            if (i == verts.length - 1) {
                p2 = verts[0];
            }
            else {
                p2 = verts[i + 1];
            }

            if (this.isClockwise(p1, p2, p) != iniResult) {
                return false;
            }
        }

        return true;
    }

另一种来自googleInterview的实现:
https://yuanhsh.iteye.com/blog/2222040

    // 是否顺时针
    private static isClockwise(a: Point, b: Point, c: Point): boolean {
        return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y) < 0;
    }
    private static isClockwiseMargin(a: Point, b: Point, c: Point): boolean {
        return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y) <= 0;
    }
    private static inConvexPolygon(p: Point, verts: Point[]): boolean {
        if (verts.length < 3) {
            return false;
        }

        if (this.isClockwise(verts[0], p, verts[1])) return false;
        if (this.isClockwise(verts[verts.length - 1], p, verts[0])) return false;

        let i = 2, j = verts.length - 1;
        let line = -1;
        while (i <= j) {
            let mid = (i + j) >> 1;
            if (this.isClockwiseMargin(verts[0], p, verts[mid])) {
                line = mid;
                j = mid - 1;
            }
            else {
                i = mid + 1;
            }
        }
        return this.isClockwise(verts[line], p, verts[line - 1]);
    }