【leetcode】Max Points on a Line

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()