LeetCode-149.Max Points on a Line

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

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

开始除了暴力破解没想到什么解决方案，暴力破解就是随机取两个点，判断其他的点在这两个点构成的直线上。代码比较繁琐就不实现了。

后来参考http://blog.csdn.net/linhuanmars/article/details/21060933

直接判断每一个点和其它点构成的斜率，然后统计同斜率的个数就好。

存在几个陷阱

1、当x1==x2，即斜率的无穷大时的处理。代码使用int.MaxValue，我觉得存在点问题：如果p1(1,-1)，p2(2,int.MaxValue-1),那么这个两个点也是的斜率k就是int.MaxValue。比如输入是[[1,-1],[1,0],[2,2147483646]]这三个点，正确输出应该是2，但是当前解决方案却输出为3。

同时还发现一个问题，提交参考博文中的代码中，用以上测试用例的Expected answer=2，但是也能通过



而使用我的C#代码提交，用以上测试用例的Expected
answer=3，也能通过。同时其它语言的Expected answer都是2，有鬼...



2、当点重合时

3、当全部点重合时，hashtable为空的判断

/**
* 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)
{
int n = points.Length, max = 2, duplicate;
if (n < 2)
return n;
double k;
Hashtable table = new Hashtable();
for (int i = 0; i < n; i++)
{
duplicate = 0;

for (int j = 0; j < n; j++)
{
if (i != j)
{
if (points[i].x == points[j].x && points[i].y == points[j].y)
duplicate++;
else
{
if (points[i].x == points[j].x)
k = int.MaxValue;
else
k = (double)(points[i].y - points[j].y) / (points[i].x - points[j].x);
if (table.Contains(k))
table[k] = (int)table[k] + 1;
else
table.Add(k, 2);
}
}
}
if (table.Count == 0)
return n;
foreach (int val in table.Values)
max = Math.Max(max, val + duplicate);
table.Clear();
}
return max;
}
}

另外，根据参考的代码，循环变量i,j的取值不一样，也需要消化一下

public int MaxPoints(Point[] points)
{
int n = points.Length, max = 2, duplicate;
if (n == 0)
return 0;
double k;
Hashtable table = new Hashtable();
for (int i = 0; i < n-1; i++)
{
duplicate = 0;
for (int j = i+1; j < n; j++)
{
if (points[i].x == points[j].x && points[i].y == points[j].y)
duplicate++;
else
{
if (points[i].x == points[j].x)
k = int.MaxValue;
else
k = (double)(points[i].y - points[j].y) / (points[i].x - points[j].x);
if (table.Contains(k))
table[k] = (int)table[k] + 1;
else
table.Add(k, 2);
}
}
if (table.Count == 0)
return n;
foreach (int val in table.Values)
max = Math.Max(max, val + duplicate);
table.Clear();
}
return max;
}