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;
}
};
