找到最快的方式2点之间的距离

本文介绍了找到最快的方式2点之间的距离的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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)); }