LeetCode：149_Max Points on a line | 寻找一条直线上最多点的数量 | Hard

题目：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.

这道题需要稍微转变一下思路，用斜率来实现，试想找在同一条直线上的点，怎么判断在一条直线上，唯一的方式也只有斜率可以完成，我开始没想到，后来看网友的思路才想到的，下面是简单的实现：
其中有一点小技巧，利用map<double, int>来存储具有相同斜率值的的点的数量，简洁高效。

/* 
    Definition for a point
*/
// struct Point {
//     int x;
//     int y;
//     Point():x(0),y(0) {}
//     Point (int a, int b):x(0), y(0) {}
// };

int maxPoints(vector<Point> &points) {
    if (points.empty())
        return 0;
    if (points.size() <= 2) 
        return points.size();

    int numPoints = points.size();
    map<double, int> pmap;    //存储斜率-点数对应值
    int numMaxPoints = 0;

    for (int i = 0; i != numPoints - 1; ++i) {
        int numSamePoints = 0, numVerPoints = 0;  //针对每个点分别做处理
        
        pmap.clear();
        
        for (int j = i + 1; j != numPoints; ++j) {
            if(points[i].x != points[j].x) {
                double slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);
                if (pmap.find(slope) != pmap.end())
                    ++ pmap[slope];   //具有相同斜率值的点数累加
                else 
                    pmap[slope] = 1;
            }
            else if (points[i].y == points[j].y) 
                numSamePoints ++;   //重合的点
            else
                numVerPoints ++;    //垂直的点

        }
        map<double, int>::iterator it = pmap.begin();
        for (; it != pmap.end(); ++ it) {
            if (it->second > numVerPoints)
                numVerPoints = it->second;
        }
        if (numVerPoints + numSamePoints > numMaxPoints) 
            numMaxPoints = numVerPoints + numSamePoints;
    }
    return numMaxPoints + 1;
}