149. Max Points on a Line

思路：这题的思路很简单，就是统计由每个函数上的点的数量，问题在于：
1.有重复点
2.对于垂直于x轴的线怎么处理
3.如果用double,那么对于斜率非常接近怎么处理
特殊用例：[0,MAX_INT-1,MAX_INT-2]
discussion里看到的大牛思路：
y0/x0 = (y2-y1)/(x2-x1) = (y3-y2)/(x3-x2) = a
b = y1-a*x1
利用辗转相除法，找到a,b

class Solution {
public:
    int getGCD(int a,int b){
        if(b == 0)return a;
        return getGCD(b,a%b);
    }
    int maxPoints(vector<Point>& points) {
        int n = points.size();
        if(n<2) return n;
        int result = 0;
        for(int i=0;i<n;i++)
        {
            map<pair<int,int>,int> lines;
            int v = 0;//垂直
            int over = 0;//重合
            int tempmax = 0;//最大
            for(int j =i+1;j<n;j++)
            {
                if(points[j].x == points[i].x && points[j].y == points[i].y){
                    over++;
                    continue;
                }
                if(points[j].x == points[i].x)
                {
                    v++;
                    tempmax = max(tempmax,v);
                    continue;
                }
                int a = points[j].x - points[i].x;
                int b = points[j].y - points[i].y;
                int gcd = getGCD(a,b);
                a /=gcd;
                b /=gcd;
                lines[make_pair(a,b)]++;
                tempmax = max(tempmax,lines[make_pair(a,b)]);
                }
             result = max(result,over+tempmax+1);
            }
        return result;
        }
};