# Anyone could find out the bugs in this algorithm please ? Help ! crazing for the result of OJ is WA . T__T

• My algorithm is below , I think its thought is simple , it depends on :

When 2 vector (v1(x1,y1) , v2(x2,y2) ) is parallel , x1/x2 === y1/y2 => x1y2 === y1x2 .

so , I list all vectors with given points , if a point (p(x0,y0)) is on a line (decided by p1(x1,y1),p2(x2,y2) , vector is v(x2-x1,y2-y1)) , the point and one end point (such as p1) of the line will make up a vector v0(x0-x1,y0-y1) , v0 is parallel with v1 .
so , (x0 - x1) * (y2 - y1) === (y0 - y1)*(x2 - x1 ) , so the set of v will add one point p
I just count points of all vector set and select max points num .

but in one case input is :

[(0,9),(138,429),(115,359),(115,359),(-30,-102),(230,709),(-150,-686),(-135,-613),(-60,-248),(-161,-481),(207,639),(23,79),(-230,-691),(-115,-341),(92,289),(60,336),(-105,-467),(135,701),(-90,-394),(-184,-551),(150,774)]

my result is

19

12

I cant find the error or bug in my program , any can help me ? Thanks a billion !

``````class Solution {
public:
int maxPoints(vector<Point> &points) {
//find them on a same line
if(points.size() <= 2)
{
return points.size();
}
std::vector<std::pair<Point,int> >	mp;
int maxn = 2;
for(int i = 0;i < points.size(); ++i)
{

mp.clear();
for(int j = i+1; j < points.size(); ++j)
{
Point pt(points[j].x-points[i].x , points[j].y - points[i].y);
std::vector<std::pair<Point,int> >::iterator it = mp.begin();
bool bFind = false;
while(it != mp.end())
{
if(pt.x*(it->first.y) == pt.y*(it->first.x))
{
it->second++;
if(maxn < it->second)
{
maxn = it->second;
}
bFind = true;
}
it++;
}
if(false == bFind)
{
mp.push_back(std::make_pair(pt,2));
}
}
}
return maxn;
}
};``````

• Your vector v(x,y) collects all line segments with slope y/x. But those segments are not necessarily in the same line, as a line is decided by its slope and y-intersect.

You can use v( (y1-y2)/(x1-x2) , (x1y2-x2y1)/(x1-x2) ) to represent the line. That should give you the right answer. Another solution is to use Ax+By+C=0 to represent the line. Each point (x0,y0) that satisfies Ax0+By0+C=0 is in this line.

• Yeap ! you are great ! thanks a billion !!

• In fact , the alogrithm collects all line , but a bug in it .
the bug is when two points (points[i],points[j]) is same , it will generate a line that past points[i] and slope = INF (0,0) .
and this line parallel with any vector (points[i],points[k]) , so if meet this , the algorithm is invalid . just except this case will be right .
in the sample case ,
(115,359),(115,359) is same ,

• Collecting all line that past p[i] and slope is (p[i]->p[j]) .
Should be carefully dealing with the slope is 0 .

• Thanks for your update! I forgot to mention that additional effort is needed for handling duplicate points and vertical lines (whose slope is INF).

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