【Leetcode】Max Points on a Line

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

每次固定一个点，然后遍历所有其它点，记录每种斜率出现的次数。需要考虑两种特殊情况：斜率不存在和点与固定点重合。

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
class Solution {
public:
    int maxPoints(vector<Point> &points) {
        int n = points.size();
        if (n <= 2) {
            return n;
        }
        int ans = 2;
        for (int i = 0; i < n - 1; ++i) {
            Point o = points[i];
            unordered_map<double, int> k_cnt;    //斜率出现计数
            int v_cnt = 0;        //垂直线上点计数（含重合点）
            int dup_cnt = 0;    //与固定点重合的点计数
            for (int j = i + 1; j < n; ++j) {
                Point p = points[j];
                if (p.x == o.x) {
                    ++v_cnt;    //垂直线上的点
                    if (p.y == o.y) {
                        ++dup_cnt;    //与固定点重合的点
                    }
                }
                else {
                    double k = ((double)p.y - o.y) / (p.x - o.x);
                    if (k_cnt.find(k) != k_cnt.end()) {
                        ++k_cnt[k];
                    }
                    else {
                        k_cnt[k] = 1;
                    }
                }
            }
            for (auto &c : k_cnt) {
                if (c.second + dup_cnt + 1 > ans) {
                    ans = c.second + dup_cnt + 1;
                }
            }
            if (v_cnt + 1 > ans) {
                ans = v_cnt + 1;
            }
        }
        return ans;
    }
};