149. Max Points on a Line (Array; Greedy)

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

思路：对于某一点来说，在经过该点的直线中选取节点数量最多的直线；对于全局来说，必定是某个局部点满足条件的直线之一=>局部最优解也是全局最优解=>贪心法。

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) {
        float k;
        int counter = 0,ret = 0,tmpMax = 0;
        unordered_map< float,int > slopes;
        for(int i = 0; i < points.size(); i++){//按点遍历，那么计算斜率就能确定一条直线，因为已有一个点的信息，
            for(int j = 0; j < points.size(); j++){
                if(points[j].x==points[i].x&&points[j].y==points[i].y){//重合的点
                    counter++;
                    continue;
                }
                k = (float)(points[j].y-points[i].y)/(points[j].x-points[i].x);
                slopes[k]++;
            }
            for(unordered_map< float,int >::iterator it=slopes.begin(); it!=slopes.end();it++){
                tmpMax =max(tmpMax, it->second);
            }
            ret = max(tmpMax+counter, ret);
            counter = 0;
            tmpMax = 0;
            slopes.clear();
        }
        return ret;
    }
};