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。
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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 precisionpublic
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;} |
[CC150] Find a line passing the most number of points,布布扣,bubuko.com
[CC150] Find a line passing the most number of points
原文地址：http://www.cnblogs.com/Antech/p/3756825.html
