[LeetCode]149. Max Points on a Line

题目描述

思路

对每个点，将所有可能的直线存储在map中，并记录直线上对应的点，每次返回点最多的个数
踩的坑：
1. 开始考虑存直线需要 y = k * x + b， 即需存储k， b两个值，后来想想，通过同一点的直线，只需要k即可确定，相同的k值代表相同的直线
2. k值存在为无穷大的问题，即直线与y轴垂直的时候。
3. k值精度问题，如果保存的是k值，当数据过大时候k值会直接相同，因此选择保存的是类似分数的形式，对diffY和diffX求最大公约数，之后保存他们对应的最简分数，可以较为准确的区分

代码

#include <iostream>#include <vector>#include <unordered_map>#include <map>#include <cmath>#include <algorithm>using namespace std;struct Point {    int x;    int y;    Point() : x(0), y(0) {}    Point(int a, int b) : x(a), y(b) {}};class Solution {public:    int GCD(int a, int b) {        while (b != 0) {            int t = b;            b = a % b;            a = t;        }        return a;    }    int maxPoint(vector<Point>& points) {        if (points.size() <= 2)            return points.size();        int res = 0;        for (int i = 0; i < points.size() - 1; ++i) {            map<pair<int, int>, int> pointCount;            int vertical = 0, sameCount = 1;            for (int j = i + 1; j < points.size(); ++j) {                if (points[j].x == points[i].x && points[j].y != points[i].y)                    ++vertical;                else if (points[j].x == points[i].x && points[j].y == points[i].y)                    ++sameCount;                else {                    //double k = (float)(points[j].y - points[i].y) / (float)(points[j].x - points[i].x);                    //long double k = atan((long double)((long double)(points[j].y - points[i].y) / points[j].x - points[i].x));                    //cout << k << endl;                    int diffY = points[j].y - points[i].y, diffX = points[j].x - points[i].x;                    int gcd = GCD(diffY, diffX);                    diffY /= gcd, diffX /= gcd;                    ++pointCount[make_pair(diffX, diffY)];                }            }            res = max(res, vertical + sameCount);            for (auto &p : pointCount) {                res = max(res, p.second + sameCount);            }        }        return res;    }};int main() {    vector<Point> points = { Point(0, 0), Point(94911151, 94911150), Point(94911152, 94911151) };    Solution s;    cout << s.maxPoint(points) << endl;    system("pause");}