leetcode-[3]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

思路：最多的点，必然是点连成线时，所有斜率相同的最多的组合情况；

　　   那么如果不在同一直线点的组合也可能斜率相同，找其中一点与其它点连即可。

#include <iostream>
#include <vector>
#include <map>

using namespace std;

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) {
        if (points.size() < 2)
            return points.size();

        int result = 0;
        for (int i = 0; i < points.size(); i++) {
            map<pair<int, int>, int> line;

            int overlap = 0, vertical = 0, curMax = 0;
            for (int j = i + 1; j < points.size(); j++)    {
                if((points[i].x == points[j].x) &&
                   (points[i].y == points[j].y)) {
                    overlap ++;
                    continue;
                }
                else if (points[i].x == points[j].x) {
                    vertical ++;
                }
                else {
                    int dx = points[i].x - points[j].x;
                    int dy = points[i].y - points[j].y;    

                    int gcd = GCD(dx, dy);

                    dx /= gcd;
                    dy /= gcd;

                    line[make_pair(dx, dy)]++;
                    curMax = max(line[make_pair(dx, dy)], curMax);
                }
                curMax = max(vertical, curMax);
            }
            result = max(result, curMax+overlap+1);
        }
        return result;
    }

    int GCD(int a, int b) {
        if (b == 0)
            return a;
        return GCD(b, a%b);
    }
};

int main() {
    vector<Point> points;

    points.push_back(Point(3, 4));
    points.push_back(Point(3, 4));
    points.push_back(Point(3, 4));
    Solution *solution = new Solution();
    cout << solution->maxPoints(points) << endl;

    // (3,10),(0,2),(0,2),(3,10)
    vector<Point> points2;
    points2.push_back(Point(3, 10));
    points2.push_back(Point(0, 2));
    points2.push_back(Point(0, 2));
    points2.push_back(Point(3, 10));
    cout << solution->maxPoints(points2) << endl;

    return 0;
}