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。
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
向量叉积
二维下叉积公式
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]);
}