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

题目：Given n points on a 2D plane, find the maximum number 
of points that lie on the same straight line.（给的点是用vector保
存）/**
* Definition for a point.
* struct Point {
* int x;
* int y;
* Point() : x(0), y(0) {}
* Point(int a, int b) : x(a), y(b) {}
* };
*/思路：
构建双重for循环，相当于起始点与终点
一个起始点和斜率可以确定一条直线，用map<double, int>来保存对应的斜率与该直线上出现的点数的值
三种情况：重合点（用dup来保存），斜率不存在的垂直点（用vertical来保存），斜率可算的正常情况class Solution {
public:
int maxPoints(vector<Point> &points) {
int size = points.size();
if(size < 0)return -1;
if(size == 0)return 0;
if(size == 1)return 1;
map<double, int> mymap;
int res = 0; //the final result
for(int i = 0; i < size; ++i){
int vertical = 0; //垂直点
int dup = 0; //重合点duplicate
int curmax = 1;
for(int j = 0; j < size; ++j){
if(i != j){
double delt_x = points[i].x - points[j].x; //delt_* must be declared double
double delt_y = points[i].y - points[j].y;
if((int)delt_y == 0 && (int)delt_x == 0){ //重合点
dup++;
}else if(delt_x == 0){ //垂直点
if(vertical == 0)vertical = 2;
else vertical++;
curmax = max(vertical, curmax);
}else{ //common situation
double k = delt_y / delt_x;
if(mymap[k] == 0)mymap[k] = 2;
else mymap[k]++;
curmax = max(mymap[k], curmax);
}
}
}
res = max(res, curmax+dup);
}
return res;
}
};