【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()