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Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
/** * 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 n = points.size(), i, j, ans = 0; if(1 == n) return 1; for(i = 0; i < n; i++) { map<double, int> m; map<int, int> x, y; int samePoint = 0; for(j = 0; j < n; j++) { int deltaX = points[i].x - points[j].x; int deltaY = points[i].y - points[j].y; if(0 == deltaX && 0 == deltaY) { samePoint++; } else if(0 == deltaX) { if(x.find(points[i].x) == x.end()) x[points[i].x] = 1; else x[points[i].x]++; } else if(0 == deltaY) { if(y.find(points[i].y) == y.end()) y[points[i].y] = 1; else y[points[i].y]++; } else { double t = 1.0 * deltaX / deltaY; if(m.find(t) == m.end()) m[t] = 1; else m[t]++; } } ans = max(ans, samePoint); for(map<double, int>::iterator it = m.begin(); it != m.end(); it++) ans = max(ans, it->second+samePoint); for(map<int, int>::iterator it = x.begin(); it != x.end(); it++) ans = max(ans, it->second+samePoint); for(map<int, int>::iterator it = y.begin(); it != y.end(); it++) ans = max(ans, it->second+samePoint); } return ans; } };
149. Max Points on a Line *HARD* 求点集中在一条直线上的最多点数
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原文地址:http://www.cnblogs.com/argenbarbie/p/5791330.html