9.7数学与概率（四）——在二维平面上，有一些点，请找出经过点数最多的那条线

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 * 功能：在二维平面上，有一些点，请找出经过点数最多的那条线。

/** * 思路：在任意两点之间画一条无线长的直线，用散列表追踪那条直线出现的次数最多。时间复杂度O(N*N) * 注意： * 1）用斜率和y轴截距来确定是否是同一条直线。 * 2）浮点数不一定能用二进制数准确表示，因此检查两个浮点数的差值是否在某个极小值（epsilon）内。 * 3）对于散列表而言，斜率相等，未必散列值相同。因此，将斜率减去一个极小值，并以得到的结果flooredSlope作为散列键。 * 4）取得所有可能相等的直线，搜索三个位置：flooredSlope,flooredSlope-epsilon,flooredSlope+epsilon。 * @param points * @return */public static MyLine1 findBestLine(GraphPoint[] points){MyLine1 bestLine=null;int bestCount=0;HashMap<Double,ArrayList<MyLine1>> lineBySlope=new HashMap<Double, ArrayList<MyLine1>>();for(int i=0;i<points.length-1;i++){for(int j=i+1;j<points.length;j++){MyLine1 line=new MyLine1(points[i],points[j]);insertLine(lineBySlope,line);int count=countEquivalentLines(lineBySlope,line);if(count>bestCount){bestCount=count;bestLine=line;}}}return bestLine;}private static int countEquivalentLines(HashMap<Double, ArrayList<MyLine1>> lineBySlope, MyLine1 line) {double key=line.floorToNearestEpsilon(line.slope);double eps=line.epsilon;int count=countEquivalentLines(lineBySlope.get(key), line)+countEquivalentLines(lineBySlope.get(key-eps), line)+countEquivalentLines(lineBySlope.get(key+eps), line);return count;}public static int countEquivalentLines(ArrayList<MyLine1> lines,MyLine1 line){if(lines==null)return 0;int count=0;for(MyLine1 paralleLine:lines){if(paralleLine==line)count++;}return count;}private static void insertLine(HashMap<Double, ArrayList<MyLine1>> lineBySlope, MyLine1 line) {ArrayList<MyLine1> lines=null;double key=line.floorToNearestEpsilon(line.slope);if(!lineBySlope.containsKey(key)){lines=new ArrayList<MyLine1>();lineBySlope.put(key, lines);}else{lines=lineBySlope.get(key);}lines.add(line);//注意此处添加的用法}class MyLine1{public static double epsilon=0.0001;public double slope,intercept;public boolean infiniteSlope=false;public MyLine1(GraphPoint p,GraphPoint q){if(Math.abs(p.x-q.x)>epsilon){//两个点的x坐标不同slope=(p.y-q.y)/(p.x-q.x);//斜率intercept=p.y-slope*p.x;//y轴截距}else{infiniteSlope=true;intercept=p.x;//x轴截距}}public double floorToNearestEpsilon(double d){int r=(int) (d/epsilon);//使原d保留小数位后的4位（epsilon=0.0001）return ((double)r)*epsilon;}public boolean isEquivalent(MyLine1 line){if((slope==line.slope)&&(intercept==line.intercept)&&(infiniteSlope==line.infiniteSlope))return true;return false;}public boolean isEquivalent(double a,double b){return Math.abs(a-b)<epsilon;}}class GraphPoint{int x;int y;public GraphPoint(int x,int y){this.x=x;this.y=y;}}

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