leetcode ex3 找出穿过最多点的直线 Max Points on a Line

题目 https://oj.leetcode.com/problems/max-points-on-a-line/

答案与分析 http://www.aiweibang.com/yuedu/183264.html

　　　　　http://blog.csdn.net/ojshilu/article/details/21618675

　　　　　http://www.1point3acres.com/bbs/thread-14425-1-1.html

　　　　http://akalius.iteye.com/blog/161852

　　　　  http://jchu.blog.sohu.com/116744856.html

　　　　   https://oj.leetcode.com/discuss/7607/accepted-code-in-python

下面是我的解法，跟http://blog.csdn.net/ojshilu/article/details/21618675 的解法是类似的，

主要思想是：1）选去平面上任意的两个点 2) 计算剩下的点会在之前两点决定的线上会有多少个3）记录最大值

我自己试了试少量的点，算法上是正确的，但是提交到leetcode就TLE（Time Limit Exceeded）了，但是上面那个链接上的用的是c++就通过了。。。。

那就只好学习下leetcode上AC的python代码了（感觉还是国内博客上的代码好懂。。。）

这个是我自己的代码

# 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 maxPoints(self, points):

        max_num = 2

        if len(points)<3:
            return len(points)

        for a in points:
            for b in points[1:]:
                sums = 2
                if b.x == a.x and b.y == a.y:
                    continue

                for c in points:
                    if c.x !=a.x and c.x != b.x\
                    and c.y!=a.y and c.y != b.y\
                    and (c.y-a.y)*(b.x-a.x) == (b.y-a.y)*(c.x-a.x):#(c.y-a.y)/(c.x-a.x) == (b.y-a.y)/(b.x-a.x)
                        sums = sums +1

                if max_num < sums :
                    max_num =sums

        return max_num

points = []
points.append(Point(40,-20))
points.append(Point(4,-2))
points.append(Point(8,-4))
points.append(Point(50,-17))

print points

c = Solution()
a = c.maxPoints(points)

print a