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; } };