leetcode #149 in cpp

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

Solution: 

 three points,p1,p2,p3, are on the same line if and only if slope(p2) == slope(p3) where slope of pi is defined as (pi.y - p1.y)/
(pi.x - p1.x). 

Think about it in the other way. If p2 has the same slope as p3 does with respect to p, then p, p3, p2 would be on the same
line. 

algorithm:

for each point p

create a map collecting slope with respect to p

 for other point pi

calculate the slope(pi) with respect to p;

    increase map[slope(pi)] ( map[slope(pi)] represents the total number of points with slope(pi) with respect to p )

find the max number in the map. 

Code:

/**
* Definition for a point.
* struct Point {
* int x;
* int y;
* Point() : x(0), y(0) {}
* Point(int a, int b) : x(a), y(b) {}
* };
*/
class Solution {
public:
int maxPoints(vector<Point>& points) {
int maxPoint = 0;
int x;
int y;
float slope;
map<float,int> slopes;

for(int i = 0; i < points.size(); i ++){
slopes.clear(); //collect slopes of point i+1...n with respect to points[i]
x = points[i].x;
y = points[i].y;
int temp_max = 1;
for(int j = i + 1; j < points.size(); j ++){
if(points[j].x == x && points[j].y == y){//if the same point
temp_max++;
continue;
}
//if not the same point, calculate slope and add 1 to the corresponding entry
slope = points[j].x - x == 0 ?INT_MAX:float(points[j].y - y )/(points[j].x -x);
if(slopes.find(slope) == slopes.end()){
slopes[slope] = 1;
}else{
slopes[slope] ++;
}
}
maxPoint = max(maxPoint, temp_max); //compare num of points[i] in the input first.
//this is because if the input contains only points[i]
// we wont go into the loop to find the max point
for(auto it = slopes.begin(); it!=slopes.end(); ++it){
maxPoint = max(maxPoint, it->second + temp_max );
}
}

return maxPoint;
}

};