Max Points on a Line

描述

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

分析

暴力枚举法。两点决定一条直线，n个点两两组合，可以得到$\dfrac{1}{2}n(n+1)$条直线，对每一条直线，判断n个点是否在该直线上，从而可以得到这条直线上的点的个数，选择最大的那条直线返回。复杂度O(n^3)。

上面的暴力枚举法以“边”为中心，再看另一种暴力枚举法，以每个“点”为中心，然后遍历剩余点，找到所有的斜率，如果斜率相同，那么一定共线对每个点，用一个哈希表，key 为斜率，value 为该直线上的点数，计算出哈希表后，取最大值，并更新全局最大值，最后就是结果。时间复杂度O(n^2)，空间复杂度O(n)。

以边为中心

Java

// Max Points on a Line
// 暴力枚举法，以边为中心，时间复杂度O(n^3)，空间复杂度O(1)
public class Solution {
    public int maxPoints(Point[] points) {
        if (points.length < 3) return points.length;
        int result = 0;

        for (int i = 0; i < points.length - 1; i++) {
            for (int j = i + 1; j < points.length; j++) {
                int sign = 0;
                int a = 0, b = 0, c = 0;
                if (points[i].x == points[j].x) sign = 1;
                else {
                    a = points[j].x - points[i].x;
                    b = points[j].y - points[i].y;
                    c = a * points[i].y - b * points[i].x;
                }
                int count = 0;
                for (int k = 0; k < points.length; k++) {
                    if ((0 == sign && a * points[k].y == c +  b * points[k].x) ||
                            (1 == sign&&points[k].x == points[j].x))
                        count++;
                }
                if (count > result) result = count;
            }
        }
        return result;
    }
}

C++

// Max Points on a Line
// 暴力枚举法，以边为中心，时间复杂度O(n^3)，空间复杂度O(1)
class Solution {
public:
    int maxPoints(vector<Point> &points) {
        if (points.size() < 3) return points.size();
        int result = 0;

        for (int i = 0; i < points.size() - 1; i++) {
            for (int j = i + 1; j < points.size(); j++) {
                int sign = 0;
                int a, b, c;
                if (points[i].x == points[j].x) sign = 1;
                else {
                    a = points[j].x - points[i].x;
                    b = points[j].y - points[i].y;
                    c = a * points[i].y - b * points[i].x;
                }
                int count = 0;
                for (int k = 0; k < points.size(); k++) {
                    if ((0 == sign && a * points[k].y == c +  b * points[k].x) ||
                        (1 == sign&&points[k].x == points[j].x))
                        count++;
                }
                if (count > result) result = count;
            }
        }
        return result;
    }
};

以点为中心

Java

// Max Points on a Line
// 暴力枚举，以点为中心，时间复杂度O(n^2)，空间复杂度O(n^2)
public class Solution {
    public int maxPoints(Point[] points) {
        if (points.length < 3) return points.length;
        int result = 0;

        HashMap<Double, Integer> slope_count = new HashMap<>();
        for (int i = 0; i < points.length-1; i++) {
            slope_count.clear();
            int samePointNum = 0; // 与i重合的点
            int point_max = 1;    // 和i共线的最大点数

            for (int j = i + 1; j < points.length; j++) {
                final double slope; // 斜率
                if (points[i].x == points[j].x) {
                    slope = Double.POSITIVE_INFINITY;
                    if (points[i].y == points[j].y) {
                        ++ samePointNum;
                        continue;
                    }
                } else {
                    if (points[i].y == points[j].y) {
                        // 0.0 and -0.0 is the same
                        slope = 0.0;
                    } else {
                        slope = 1.0 * (points[i].y - points[j].y) /
                                (points[i].x - points[j].x);
                    }
                }

                int count = 0;
                if (slope_count.containsKey(slope)) {
                    final int tmp = slope_count.get(slope);
                    slope_count.put(slope, tmp + 1);
                    count = tmp + 1;
                } else {
                    count = 2;
                    slope_count.put(slope, 2);
                }

                if (point_max < count) point_max = count;
            }
            result = Math.max(result, point_max + samePointNum);
        }
        return result;
    }
}

C++

// Max Points on a Line
// 暴力枚举，以点为中心，时间复杂度O(n^2)，空间复杂度O(n^2)
class Solution {
public:
    int maxPoints(vector<Point> &points) {
        if (points.size() < 3) return points.size();
        int result = 0;

        unordered_map<double, int> slope_count;
        for (int i = 0; i < points.size()-1; i++) {
            slope_count.clear();
            int samePointNum = 0; // 与i重合的点
            int point_max = 1;    // 和i共线的最大点数

            for (int j = i + 1; j < points.size(); j++) {
                double slope; // 斜率
                if (points[i].x == points[j].x) {
                    slope = std::numeric_limits<double>::infinity();
                    if (points[i].y == points[j].y) {
                        ++ samePointNum;
                        continue;
                    }
                } else {
                    if (points[i].y == points[j].y) {
                        // 0.0 and -0.0 is the same
                        slope = 0.0;
                    } else {
                        slope = 1.0 * (points[i].y - points[j].y) /
                                (points[i].x - points[j].x);
                    }
                }

                int count = 0;
                if (slope_count.find(slope) != slope_count.end())
                    count = ++slope_count[slope];
                else {
                    count = 2;
                    slope_count[slope] = 2;
                }

                if (point_max < count) point_max = count;
            }
            result = max(result, point_max + samePointNum);
        }
        return result;
    }
};