[LeetCode] 149. Max Points on a Line 共线点个数

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

给一个由n个点组成的2D平面，找出最多的同在一条直线上的点的个数。

共线点的条件是斜率一样，corn case：点相同；x坐标相同。

Java：

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public class Solution {     public int maxPoints(Point[] points) {         int res = 0;         for (int i = 0; i < points.length; ++i) {             Map<Map<Integer, Integer>, Integer> m = new HashMap<>();             int duplicate = 1;             for (int j = i + 1; j < points.length; ++j) {                 if (points[i].x == points[j].x && points[i].y == points[j].y) {                     ++duplicate; continue;                 }                 int dx = points[j].x - points[i].x;                 int dy = points[j].y - points[i].y;                 int d = gcd(dx, dy);                 Map<Integer, Integer> t = new HashMap<>();                 t.put(dx / d, dy / d);                 m.put(t, m.getOrDefault(t, 0) + 1);             }             res = Math.max(res, duplicate);             for (Map.Entry<Map<Integer, Integer>, Integer> e : m.entrySet()) {                 res = Math.max(res, e.getValue() + duplicate);             }         }         return res;     }     public int gcd(int a, int b) {         return (b == 0) ? a : gcd(b, a % b);     } }

Python：

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class Point:     def __init__(self, a=0, b=0):         self.x = a         self.y = b   class Solution(object):     def maxPoints(self, points):         """         :type points: List[Point]         :rtype: int         """         max_points = 0         for i, start in enumerate(points):             slope_count, same = collections.defaultdict(int), 1             for j in xrange(i + 1, len(points)):                 end = points[j]                 if start.x == end.x and start.y == end.y:                     same += 1                 else:                     slope = float("inf")                     if start.x - end.x != 0:                         slope = (start.y - end.y) * 1.0 / (start.x - end.x)                     slope_count[slope] += 1                           current_max = same                        for slope in slope_count:                 current_max = max(current_max, slope_count[slope] + same)                               max_points = max(max_points, current_max)                       return max_points

C++：

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class Solution { public:     int maxPoints(vector<Point>& points) {         int res = 0;         for (int i = 0; i < points.size(); ++i) {             map<pair<int, int>, int> m;             int duplicate = 1;             for (int j = i + 1; j < points.size(); ++j) {                 if (points[i].x == points[j].x && points[i].y == points[j].y) {                     ++duplicate; continue;                 }                 int dx = points[j].x - points[i].x;                 int dy = points[j].y - points[i].y;                 int d = gcd(dx, dy);                 ++m[{dx / d, dy / d}];             }             res = max(res, duplicate);             for (auto it = m.begin(); it != m.end(); ++it) {                 res = max(res, it->second + duplicate);             }         }         return res;     }     int gcd(int a, int b) {         return (b == 0) ? a : gcd(b, a % b);     } };

　　　　

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