Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
This question is a very typical one using reversed thinking.
Two different way to start:
- First one is pretty straight forward, calculate there are how many different lines on this 2D panel and count the points number on each line to get the max one.
- Second one is a little bit difficult to consider. We take advantage of the principal: if we have a fixed point and a slope, then we can exactly define a line. So for each point, we keep a hashmap, using the line slope as its key, and the value is the number of points which fell onto the line except the duplicate points.
/**
* Definition for a point.
* struct Point {
* int x;
* int y;
* Point() : x(0), y(0) {}
* Point(int a, int b) : x(a), y(b) {}
* };
*/
int maxPoints(vector<Point>& points){
if(points.size() <= 2) return points.size();
int maxNum = 0;
for(int i = 0; i < points.size(); i++){
unordered_map<double,int> map;
map[INT_MAX] = 0;
int duplicate = 1, resMax = 0;
for(int j = i+1; j < points.size(); j++){
if(points[j].x == points[i].x && points[j].y == points[i].y){
duplicate++;
}else{
double slope = (points[j].x == points[i].x) ? INT_MAX : (double)(points[i].y-points[j].y)/(points[i].x-points[j].x);
map[slope]++;
resMax = max(resMax,map[slope]);
}
}
maxNum = max(maxNum,resMax+duplicate);
}
return maxNum;
}
