[LintCode] Max Points on a Line 共线点个数

最新推荐文章于 2021-05-23 18:48:53 发布

转载最新推荐文章于 2021-05-23 18:48:53 发布 · 104 阅读

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原文链接：https://yq.aliyun.com/articles/336190

本文提供了解决二维平面上共线点最大数量问题的三种算法实现方案，包括使用unordered_map存储斜率、利用map简化斜率计算及直接通过坐标判断是否共线的方法。

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

Have you met this question in a real interview?

Example

Given 4 points: (1,2), (3,6), (0,0), (1,3).

The maximum number is 3.

LeetCode上的原题，请参见我之前的博客Max Points on a Line。

解法一：

class Solution {
public:
    /**
     * @param points an array of point
     * @return an integer
     */
    int maxPoints(vector<Point>& points) {
        int res = 0, n = points.size();
        for (int i = 0; i < n; ++i) {
            int duplicate = 1;
            unordered_map<double, int> m;
            for (int j = i + 1; j < n; ++j) {
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    ++duplicate;
                } else if (points[i].x == points[j].x) {
                    ++m[INT_MAX];
                } else {
                    double slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);
                    ++m[slope];
                }
            }
            res = max(res, duplicate);
            for (auto it : m) {
                res = max(res, it.second + duplicate);
            }
        }
        return res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param points an array of point
     * @return an integer
     */
    int maxPoints(vector<Point>& points) {
        int res = 0, n = points.size();
        for (int i = 0; i < n; ++i) {
            int duplicate = 1;
            map<pair<int, int>, int> m;
            for (int j = i + 1; j < n; ++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) {
                res = max(res, it.second + duplicate);
            }
        }
        return res;
    }
    int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }
};

解法三：

class Solution {
public:
    /**
     * @param points an array of point
     * @return an integer
     */
    int maxPoints(vector<Point>& points) {
        int res = 1;
        for (int i = 0; i < points.size(); ++i) {
            int repeat = 1;
            for (int j = i + 1; j < points.size(); ++j) {
                int cnt = 0;
                int x1 = points[i].x, y1 = points[i].y;
                int x2 = points[j].x, y2 = points[j].y;
                if (x1 == x2 && y1 == y2) {++repeat; continue;}
                for (int k = 0; k < points.size(); ++k) {
                    int x3 = points[k].x, y3 = points[k].y;
                    if (x1 * y2 + x2 * y3 + x3 * y1 - x3 * y2 - x2 * y1 - x1 * y3 == 0) {
                        ++cnt;
                    }
                }
                res = max(res, cnt);
            }
            res = max(res, repeat);
        }
        return points.empty() ? 0 : res;
    }
};