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* update github.com/blevesearch/bleve v2.0.2 -> v2.0.3 * github.com/denisenkom/go-mssqldb v0.9.0 -> v0.10.0 * github.com/editorconfig/editorconfig-core-go v2.4.1 -> v2.4.2 * github.com/go-chi/cors v1.1.1 -> v1.2.0 * github.com/go-git/go-billy v5.0.0 -> v5.1.0 * github.com/go-git/go-git v5.2.0 -> v5.3.0 * github.com/go-ldap/ldap v3.2.4 -> v3.3.0 * github.com/go-redis/redis v8.6.0 -> v8.8.2 * github.com/go-sql-driver/mysql v1.5.0 -> v1.6.0 * github.com/go-swagger/go-swagger v0.26.1 -> v0.27.0 * github.com/lib/pq v1.9.0 -> v1.10.1 * github.com/mattn/go-sqlite3 v1.14.6 -> v1.14.7 * github.com/go-testfixtures/testfixtures v3.5.0 -> v3.6.0 * github.com/issue9/identicon v1.0.1 -> v1.2.0 * github.com/klauspost/compress v1.11.8 -> v1.12.1 * github.com/mgechev/revive v1.0.3 -> v1.0.6 * github.com/microcosm-cc/bluemonday v1.0.7 -> v1.0.8 * github.com/niklasfasching/go-org v1.4.0 -> v1.5.0 * github.com/olivere/elastic v7.0.22 -> v7.0.24 * github.com/pelletier/go-toml v1.8.1 -> v1.9.0 * github.com/prometheus/client_golang v1.9.0 -> v1.10.0 * github.com/xanzy/go-gitlab v0.44.0 -> v0.48.0 * github.com/yuin/goldmark v1.3.3 -> v1.3.5 * github.com/6543/go-version v1.2.4 -> v1.3.1 * do github.com/lib/pq v1.10.0 -> v1.10.1 again ...
66 lines
1.9 KiB
Go
Vendored
66 lines
1.9 KiB
Go
Vendored
// SPDX-License-Identifier: MIT
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package identicon
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var (
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// 4 个元素分别表示 cos(0),cos(90),cos(180),cos(270)
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cos = []int{1, 0, -1, 0}
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// 4 个元素分别表示 sin(0),sin(90),sin(180),sin(270)
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sin = []int{0, 1, 0, -1}
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)
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// 将 points 中的所有点,以 x,y 为原点旋转 angle 个角度。
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// angle 取值只能是 [0,1,2,3],分别表示 [0,90,180,270]
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func rotate(points []int, x, y int, angle int) {
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if angle < 0 || angle > 3 {
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panic("rotate:参数angle必须0,1,2,3三值之一")
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}
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for i := 0; i < len(points); i += 2 {
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px, py := points[i]-x, points[i+1]-y
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points[i] = px*cos[angle] - py*sin[angle] + x
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points[i+1] = px*sin[angle] + py*cos[angle] + y
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}
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}
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// 判断某个点是否在多边形之内,不包含构成多边形的线和点
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// x,y 需要判断的点坐标
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// points 组成多边形的所顶点,每两个元素表示一点顶点,其中最后一个顶点必须与第一个顶点相同。
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func pointInPolygon(x, y int, points []int) bool {
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if len(points) < 8 { // 只有2个以上的点,才能组成闭合多边形
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return false
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}
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// 大致算法如下:
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// 把整个平面以给定的测试点为原点分两部分:
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// - y>0,包含(x>0 && y==0)
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// - y<0,包含(x<0 && y==0)
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// 依次扫描每一个点,当该点与前一个点处于不同部分时(即一个在 y>0 区,一个在 y<0 区),
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// 则判断从前一点到当前点是顺时针还是逆时针(以给定的测试点为原点),如果是顺时针 r++,否则 r--。
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// 结果为:2==abs(r)。
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r := 0
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x1, y1 := points[0], points[1]
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prev := (y1 > y) || ((x1 > x) && (y1 == y))
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for i := 2; i < len(points); i += 2 {
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x2, y2 := points[i], points[i+1]
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curr := (y2 > y) || ((x2 > x) && (y2 == y))
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if curr == prev {
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x1, y1 = x2, y2
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continue
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}
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if mul := (x1-x)*(y2-y) - (x2-x)*(y1-y); mul >= 0 {
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r++
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} else if mul < 0 {
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r--
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}
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x1, y1 = x2, y2
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prev = curr
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}
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return r == 2 || r == -2
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}
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