Ale*_*erg 3 line cgpoint swift
我一直在尝试将Douglas-Peucker算法应用到我的代码中,我能够将伪代码转换为Swift,除了shortestDistanceToSegment函数.只有我能找到的Swift版本才能在这里得到解答,但我不明白它实际上是做什么的.
我需要一个函数,它获得三个点作为参数(点和线的两端)并返回CGPoint和线段之间的最短距离.关于代码做什么(和为什么)的一些解释很好但不是必要的.
Sco*_*ter 10
回答来自/sf/answers/1941595701/ w /变量重命名和添加的一些评论:
/* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment v: CGPoint, and w: CGPoint) -> CGFloat {
let pv_dx = p.x - v.x
let pv_dy = p.y - v.y
let wv_dx = w.x - v.x
let wv_dy = w.y - v.y
let dot = pv_dx * wv_dx + pv_dy * wv_dy
let len_sq = wv_dx * wv_dx + wv_dy * wv_dy
let param = dot / len_sq
var int_x, int_y: CGFloat /* intersection of normal to vw that goes through p */
if param < 0 || (v.x == w.x && v.y == w.y) {
int_x = v.x
int_y = v.y
} else if param > 1 {
int_x = w.x
int_y = w.y
} else {
int_x = v.x + param * wv_dx
int_y = v.y + param * wv_dy
}
/* Components of normal */
let dx = p.x - int_x
let dy = p.y - int_y
return sqrt(dx * dx + dy * dy)
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
1238 次 |
| 最近记录: |