题目描述：
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
思路：
这算是一道数学题，关键在于存方程，把点一一代入判断是否满足，
LeetCode Max Points on a Line
。1. 两层循环遍历节点O，存直线方程的关键因子：斜率：K, 截距：B，X：X轴截距（K为无穷时），Y:Y轴截距（K为0时）。
2. 对每个方程分别遍历每个点，找到满足最多点的方程即可。
注：在小于精度0.00001时，本解法有Bug
实现代码：
/** * Definition for a point. * public class Point { * public int x; * public int y; * public Point() { x = 0; y = 0; } * public Point(int a, int b) { x = a; y = b; } * } */public class Solution { public int MaxPoints(Point[] points) { if(points.Length == 0){ return 0; } if(points.Length < 2){ return 1; } // 1. same line functions into list var funcs = new Dictionary<string, linefunc="">(); for(var i = 0;i < points.Length; i++){ for(var j = 0; j < points.Length;j ++){ if(i != j){ var line = Line(points[i], points[j]); var k = line.ToString(); if(!funcs.ContainsKey(k)){ funcs.Add(k,line); } } } } // 2. loop each function , count how many points can fulfil var max = 0; foreach(var k in funcs.Keys){ var c = 0; for(var i = 0;i < points.Length; i++){ if(funcs[k].Test(points[i])){ c++; } } if(c > max){ max = c; } } return max; }private LineFunc Line(Point p1 , Point p2){ double deltaX = p1.x - p2.x; var k = deltaX == 0 ? int.MaxValue : ((p1.y - p2.y) / deltaX); var b = p1.y - k * p1.x; return new LineFunc(k,b,p1.x,p1.y);}public class LineFunc{ public LineFunc(double k , double b, double x0, double y0){ K = k; if(k == 0){ Y = y0; } else if(k == int.MaxValue){ X = x0; } else{ B = b; } } public bool Test(Point p){ if(K == int.MaxValue){ return p.x == X; } else if(K == 0){ return p.y == Y; } else{ var delta = p.y - (p.x * K + B); return delta < 0.00001 && delta > -0.000001; } } public double K; public double Y; public double X; public double B; public override string ToString(){ return string.Format({0}_{1}_{2}_{3}, K, B, X, Y); }}}</string,>