[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：

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：

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++：

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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