[CC150] Find a line passing the most number of points

Problem: Given a two-dimensional graph with points on it, find a line which passes the most number of points.

此题是Cracking the code 5th edition 第七章第六题，思路就是 n choose 2, 所以时间复杂度是O(n^2)，因为没有更快的办法。

此题的难点在于两点一线计算出的斜率是浮点型，不好比较equality。所以其中需要有一个精确到哪一位的概念，英文是 round to a given place value.

我认为此题书中给的解法特别傻逼，而且时间复杂度也超出了O(n^2)，故自己写了一个更好的版本。

另，关于使用自定义类用作HashMap的键值，如何重写equals()和hashCode()，下面的代码给出的很好的示范。

package chapter7;

import java.util.HashMap;

// given a two-dimensional graph with points on it,
// find a line which passes the most number of points
// Time: O(N^2), N is number of points

// The tricky part is checking the equality of slope
// which is of type double.
// My solution is floor all values to an epsilon value
// which specifies the desired precision

public class P6 {
	
	public Line findBestLine(GraphPoint[] points){
		Line bestLine = null;
		int bestCount = 0;
		HashMap<Line, Integer> lineCounts = 
				new HashMap<Line, Integer>();
		
		for(int i = 0; i < points.length; ++i){
			for(int j = i+1; j < points.length; ++j){
				Line line = new Line(points[i], points[j]);
				int currentCount;
				
				if(lineCounts.containsKey(line)){
					currentCount = lineCounts.get(line) + 1;
				}else{
					currentCount = 1;
				}
				lineCounts.put(line, currentCount);
				
				if(currentCount > bestCount){
					bestCount = currentCount;
					bestLine = line;
				}
			}
		}
		
		return bestLine;
	}
}


class Line{
	// for precision 
	// slope and intercept values are floored to epsilon
	public static double epsilon = .0001;
	
	// properties for a normal line
	public double slope;
	public double y_intercept;
	
	// properties for a verticle line
	public boolean infinite_slope = false;
	public double x_intercept;
	
	public Line(GraphPoint p1, GraphPoint p2){
		
		if(p1.x == p2.x){
			this.infinite_slope = true;
			this.x_intercept = p1.x;
			
		}else{
			this.slope = (p1.y - p2.y) / (p1.x - p2.x);
			this.y_intercept = p1.y - slope * p1.x;
		
		}
		
		// floor all properties
		this.slope = floor(this.slope);
		this.x_intercept = floor(this.x_intercept);
		this.y_intercept = floor(this.y_intercept);
	}

	public double floor(double val){
		int val2 = (int)(val / epsilon);
		return val2 * epsilon;
	}
	
	@Override
	public int hashCode(){
		if(infinite_slope){
			return (int) x_intercept;
		}else{
			return (int) (slope + y_intercept);
		}
	}
	
	@Override
	public boolean equals(Object obj){
		if(this == obj)
			return true;
		if(obj == null)
			return false;
		if(getClass() != obj.getClass())
			return false;
		
		Line other = (Line)obj;
		
		if(infinite_slope && other.infinite_slope){ // both true
			return x_intercept == other.x_intercept;
			
		}else if(infinite_slope || other.infinite_slope){ // one true, one false
			return false;
		}
		else{ // both false
			return slope == other.slope && y_intercept == other.y_intercept;
		}
	}
}


class GraphPoint{
	// assume that x and y are both floored
	// to some point
	public double x;
	public double y;
}