ios - 如何在 Swift 中获取一条线上的所有点？

Question

我正在用 Swift 编写一个程序，它需要一种方法，该方法使用两个 CGPoints 的规范并返回它们之间直线上的所有点。目前我正在试用以下方法，但它非常故障并且拒绝为某些行获得所有分数。我只是想知道是否有更有效的方法来做到这一点？

func addPointsInLine(#start: CGPoint, end: CGPoint){

    var speed = 0
    var startNo = trackPoints.count
    var endNo = startNo

    var xDiff = start.x - end.x
    var yDiff = start.y - end.y

    var xChange = 0.0
    var yChange = 0.0

    var newPointX = start.x
    var newPointY = start.y

    var ended = false

    xChange = Double(xDiff) / Double(yDiff)
    yChange = Double(yDiff) / Double(xDiff)

    /*if(xDiff > 0){
        xChange = sqrt(xChange * xChange)
    }
    if(yDiff > 0){
        yChange = sqrt(yChange * yChange)
    }*/

    println("xc \(xChange)")
    println("yc \(yChange)")

    var y = Double(start.y)
    var x = Double(start.x)

    while !ended {
        println(trackPoints.count)
        speed++
        endNo++
        if(CGPointMake(newPointX, newPointY) == end){
            ended = true
        }

        if(yChange > xChange){
            y++
            x += xChange
            trackPoints.append(TrackPoint(x: Int(x), y: Int(y)))
            if(CGFloat(Int(y)) == end.y){
                ended = true
                println("end")
                //break
            }
           /* if(yChange > 0){
                if(CGFloat(Int(y)) > end.y){
                    ended = true
                    println("end>y")
                   // break
                }
            }
            if(yChange < 0){
                if(CGFloat(Int(y)) < end.y){
                    ended = true
                    println("end<y")
                    //break
                }
            }*/
            if(xChange > 0){
                if(CGFloat(Int(x)) >= end.x){
                    ended = true
                    println("end>x")
                    //break
                }
            }
            if(xChange < 0){
                if(CGFloat(Int(x)) <= end.x){
                    ended = true
                    println("end<x")
                    //break
                }
            }
        } else {
            x++
            y += yChange
            trackPoints.append(TrackPoint(x: Int(x), y: Int(y)))
            if(CGFloat(Int(x)) == end.x){
                ended = true
                println("end")
                //break
            }
           /* if(xChange > 0){
                if(CGFloat(Int(x)) >= end.x){
                    ended = true
                    println("end>x")
                    //break
                }
            }
            if(xChange < 0){
                if(CGFloat(Int(x)) <= end.x){
                    ended = true
                    println("end<x")
                    //break
                }
            }*/
            if(yChange > 0){
                if(CGFloat(Int(y)) > end.y){
                    ended = true
                    println("end>y")
                    // break
                }
            }
            if(yChange < 0){
                if(CGFloat(Int(y)) < end.y){
                    ended = true
                    println("end<y")
                    //break
                }
            }
        }

    }

    println("finished")

    var i = startNo
    println(startNo)
    println(endNo)
    while i < endNo {
        trackPoints[i].speed = speed
        i++
    }
    println("finish2")

}

Answer 1

我认为首先找到所有点是不可能的，因为直线上有无限多个点。以连接 (0,0) 到 (1, 0) 的线为例。以下所有点以及更多点都在上述线上 (0.00001,0) (0.0000000000001,0) (0.01,0)

所以你需要限制你需要找到的点的数量，就像所有具有整数坐标的点一样。所有点相距 1 个单位，从起点等开始。

接下来您可以使用其中一个直线方程来获得分数:Equations of Line