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。

/**
 * Definition for a point.
 * 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) {
        int ret = 0;
        int subMax = 0;
        double slope;
        int sameCounter = 1;
        int zeroCounter = 0;
        map<double,int> count;
        for(int i = 0; i < points.size(); i++){
            for(int j = i+1; j < points.size(); j++){
                if(points[j].x==points[i].x && points[j].y==points[i].y) sameCounter++;
                else if(points[j].x-points[i].x == 0){
                     zeroCounter++;
                     if(zeroCounter>subMax) subMax = zeroCounter;
                }
                else{
                    slope = (double) (points[j].y-points[i].y)/(points[j].x-points[i].x);
                    count[slope]++;
                    if(count[slope]>subMax) subMax=count[slope];
                }
            }
            count.clear();
            zeroCounter = 0;
            if(subMax+sameCounter > ret) ret = subMax+sameCounter;
            subMax = 0;
            sameCounter = 1;
        }
        return ret;
    }
};