Max Points on a Line

题目描述：

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

解题思路：

1.首先由这么一个O(n^3)的方法，也就是算出每条线的方程（n^2），然后判断有多少点在每条线上（N）。这个方法肯定是可行的，只是复杂度太高
2.然后想到一个O(N)的，对每一个点，分别计算这个点和其他所有点构成的斜率，具有相同斜率最多的点所构成的直线，就是具有最多点的直线。

注意的地方：

1.重合的点
2.斜率不存在的点

 # Definition for a point

 class Point:

     def __init__(self, a=0, b=0):

         self.x = a

         self.y = b

 class Solution:

     # @param points, a list of Points

     # @return an integer

     def calcK(self,pa, pb):

         t = ((pb.y - pa.y) * 1.0) / (pb.x - pa.x)

         return t

     def maxPoints(self, points):

         l = len(points)

         res = 0

         if l <= 2:

             return l

         for i in xrange(l):

             same = 0

             k = {}

             k['inf'] = 0

             for j in xrange(l):

                 if points[j].x == points[i].x and points[j].y != points[i].y:

                     k['inf'] += 1

                 elif points[j].x == points[i].x and points[j].y == points[i].y:

                     same +=1

                 else:

                     t = self.calcK(points[j],points[i])

                     if t not in k.keys():

                         k[t] = 1

                     else:

                         k[t] += 1

             res = max(res, max(k.values())+same)

         return res

 def main():

     points = []

     points.append(Point(0,0))

     points.append(Point(1,1))

     points.append(Point(1,-1))

     #points.append(Point(0,0))

     #points.append(Point(1,1))

     #points.append(Point(0,0))

     s = Solution()

     print s.maxPoints(points)

 if __name__ == '__main__':

     main()