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java - Android/Java 如何在没有 line2d 的情况下检查矩形和线段是否相交

转载 作者:太空狗 更新时间:2023-10-29 14:28:03 27 4
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在我的 Android 游戏中,我需要检查矩形和线段之间的交点。我不能使用 line2d因为安卓不支持。我看过处理线条的类似问题并尝试修改它们但失败了。我也意识到这一点 Q/A这基本上是我想要的,但是我失败了。例如,这里是我的 Line 类代码,其中包括我在交叉点的尝试。结果是一些非交叉点返回 true,一些交叉点返回 false。编辑:Oli Charlesworth 帮助了我,这里是适用于所有 google 员工的工作代码。

   package com.example.HelloAndroid;

import android.graphics.Rect;

public class Segment {
int x1;
int y1;
int x2;
int y2;
double m;
double b;
boolean ishoriz;
boolean isvert;

public Segment(int x1s, int y1s, int x2s, int y2s) {
if (x1s > x2s) {
this.x1 = x2s;
this.x2 = x1s;
} else {
this.x1 = x1s;
this.x2 = x2s;
}
if (y1s > y2s) {
this.y1 = y2s;
this.y2 = y1s;
} else {
this.y1 = y1s;
this.y2 = y2s;
}
int ydif = y2s - y1s;
int xdif = x2s - x1s;
if (ydif == 0) {
this.ishoriz = true;
this.m = 0;
this.b = x1s;
} else if (xdif == 0) {
this.isvert = true;
} else {
this.m = (double) ydif / xdif;
double r = (double) ydif / xdif;
this.b = y1s - (r * x1s);
this.isvert = false;
this.ishoriz = false;
}
}

public final boolean intersected(Segment s, Segment s2) {
if (s.ishoriz && s2.ishoriz) {
//parallel
return false;
}

if (s.isvert && s2.isvert) {
//parallel

return false;
}

if (s.isvert) {
//x is constant see if the x is on the other line
int x = s.x1;
//add 2 for round-off error
if (s2.x1 <= x + 2 && s2.x2 + 2 >= x) {
//solve and check if y is on both segments
int y = (int) ((s.m * x) + s.b);
if(s.y1<=y+2&&s.y2+2>=y)
{
if(s2.y1<=y+2&&s2.y2+2>=y)
{
return true;}
}
}
return false;
}

if (s2.isvert) {
//x is constant see if the x is on the other line
int x = s2.x1;
//add 2 for round-off error
if (s.x1 <= x + 2 && s.x2 + 2 >= x) {
//solve and check if y is on both segments
int y = (int) ((s.m * x) + s.b);
if(s.y1<=y+2&&s.y2+2>=y)
{
if(s2.y1<=y+2&&s2.y2+2>=y)
{
return true;}
}
}
return false;
}

if (s.ishoriz) {
//y is constant see if the y is on the other line
int y = s.y1;
//add 2 for round-off error
if (s2.y1 <= y + 2 && s2.y2 + 2 >= y) {
//solve and check if x is on both segments
int x=(int) ((y-s.b)/s.m);
if(s.x1<=x+2&&s.x2+2>=x)
{
if(s2.x1<=x+2&&s2.x2+2>=x)
return true;}
return false;
}}

if (s2.ishoriz) {
//y is constant see if the y is on the other line
int y = s2.y1;
//add 2 for round-off error
if (s.y1 <= y + 2 && s.y2 + 2 >= y) {
//solve and check if x is on both segments
int x=(int) ((y-s.b)/s.m);
if(s.x1<=x+2&&s.x2+2>=x)
{
if(s2.x1<=x+2&&s2.x2+2>=x)
return true;}
}
return false;
}

if (s.m == s2.m) {
//parallel
return false;
}

// use substitution
// (s.m-s2.m)x=s2.b-s.b
int x = (int) (s.m - s2.m);
x = (int) ((s2.b - s.b) / x);
// find y
int y = (int) ((x * s.m) + s.b);
//check if the values are in between for both lines
//add 2 for round-off error
if (s.y1 <= y + 2) {
if (s.y2 + 2 >= y) {
if (s2.y1 <= y + 2) {
if (s2.y2 + 2 >= y) {
if (s.x1 <= x + 2) {
if (s.x2 + 2 >= x) {
if (s2.x1 <= x + 2) {
if (s2.x2 + 2 >= x) {
return true;
}
}
}
}
}
}
}
}
return false;
}

public final boolean intersects2(Segment s, Rect r) {
//created lines of the rect
Segment top = new Segment(r.left, r.top, r.right, r.top);
Segment left = new Segment(r.left, r.top, r.left, r.bottom);
Segment bottom = new Segment(r.left, r.bottom, r.right, r.bottom);
Segment right = new Segment(r.right, r.top, r.right, r.bottom);
boolean topp = s.intersected(s, top);
if (topp) {
return true;
}
boolean leftp = s.intersected(s, left);
if (leftp) {
return true;
}
boolean bottomp = s.intersected(s, bottom);
if (bottomp) {
return true;
}
boolean rightp = s.intersected(s, right);
if (rightp) {
return true;
} else {
return false;
}
}

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