找到最快的方式2点之间的距离 [英] Find distance between 2 points in fastest way

问题描述

这code。通过使用距离公式，的Math.sqrt计算2点之间的距离（ - X2）^ 2 +（Y - （X1 Y2）^ 2）。我的第一点具有 MMX 和 MMY 协调和第二个具有牛和 OY 协调。我的问题很简单，没​​有任何的更快办法计算吗？

This code calculates the distance between 2 points by using distance formula, Math.sqrt ( (x1 – x2)^2 + (y1 – y2) ^2). My first point has mmx and mmy coordination and second one has ox and oy coordination. My question is simple, is there any FASTER way for calculate this?

private function dist(mmx:int, mmy:int, ox:int, oy:int):Number{
  return Math.sqrt((mmx-ox)*(mmx-ox)+(mmy-oy)*(mmy-oy));
}

这是我的code，感谢您的帮助。

This is my code, Thanks for help.

public function moveIT(Xmouse, Ymouse):void{
            f = Point.distance( new Point( Xmouse, Ymouse ), new Point( mainSP.x, mainSP.y ) );// distance between mouse and instance 
            distancePro = Point.distance( pointO, new Point( mainSP.x, mainSP.y ) );// distance from start point 
            if (  f < strtSen ){ // move forward
                tt.stop(); tt.reset(); // delay timer on destination    
                mF = true;  mB = false;
                ag = Math.atan2((Ymouse - mainSP.y),(Xmouse - mainSP.x)); // move-forward angle, between mouse and instance
            }
            if (mF){ /// shoot loop
                if (f > 5){// 5 pixel
                    mainSP.x -= Math.round( (400 /f) + .5 ) * Math.cos(ag);
                    mainSP.y -= Math.round( (400 /f) + .5 ) * Math.sin(ag);
                }
                if (  distancePro > backSen ){// (backSen = max distance)
                    mF = false;         
                    tt.start();// delay timer on destination
                }
            }
            if (mB){ /// return loop
                if ( distancePro < 24 ){//  back angle re-calculation
                    agBACK = Math.atan2((y1 - mainSP.y),(x1 - mainSP.x));                   
                }
                mainSP.x += (Math.cos(agBACK) * rturnSpeed);
                mainSP.y += (Math.sin(agBACK) * rturnSpeed);
                if ( distancePro < 4 ){ // fix position to start point (x1,y1)
                    mB = false;
                    mainSP.x = x1; mainSP.y = y1;
                }
            }
        }
private function scTimer(evt:TimerEvent):void {// timer
            tt.stop();
            agBACK = Math.atan2((y1 - mainSP.y),(x1 - mainSP.x));// move-back angle between start point and instance
            mB = true;
        }

另外： pointO =新的点（X1，Y1）; 设置起点。我不能使用，因为这样的应用程序调用父类mouseX和mouseY的，所以我只能通过x和y，以我的循环。

Also: pointO = new Point(x1,y1); set start point. I can not use mouseX and mouseY because of the way that the application is called by parent class, so I can just pass x and y to my loop.

推荐答案

调用静态函数是有点贵。您可以保存开销这样做：

Calling a static function is a bit expensive. You can save that overhead by doing this:

private var sqrtFunc = Math.sqrt;

private function dist(mmx:int, mmy:int, ox:int, oy:int):Number{
    return sqrtFunc((mmx-ox)*(mmx-ox)+(mmy-oy)*(mmy-oy));
}

这篇关于找到最快的方式2点之间的距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！