[LintCode] 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.

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;
    }
};