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【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.

 

1st (7 tries)

class Solution {public:    int maxPoints(vector<Point> &points) {        unordered_map<float,int> kmap;		int maxres = 0;		for(int i = 0;i < points.size();i++) {			int same_num = 1;			kmap.clear();			kmap[INT_MIN] = 0;			for(int j = 0;j < points.size();j++) {				if (j == i)					continue;				if (points[i].x == points[j].x && points[i].y == points[j].y) {					same_num ++ ;					continue;				}				float k = points[i].x == points[j].x?INT_MAX:(float)(points[j].y - points[i].y)/(points[j].x - points[i].x);  				kmap[k]++;			}			unordered_map<float,int>::iterator iter;			for(iter = kmap.begin();iter != kmap.end();++iter) {				if(iter->second+same_num>maxres) {					maxres = iter->second+same_num;				}			}		}		return maxres;    }};

  

2nd ( 4 tires)

/** * 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 max_p = 0;        for(unsigned i = 0;i < points.size();i++) {            map<double,int> slopes;            int min_p = 0;            for(unsigned j = 0;j < points.size();j++) {                if(points[i].x == points[j].x && points[i].y == points[j].y) {                    min_p++;                    continue;                }                double k = (points[i].x - points[j].x)?( (double)(points[i].y - points[j].y) )/(points[i].x - points[j].x):INT_MAX;                if(slopes.find(k) == slopes.end())                    slopes[k] = 1;                else                    slopes[k]++;            }            if(slopes.size() == 0) {                if(min_p > max_p)                    max_p = min_p;            }            else {                for(map<double,int>::iterator iter = slopes.begin();iter != slopes.end();++iter) {                    if(iter->second + min_p > max_p)                        max_p = iter->second + min_p;                }            }        }        return max_p;    }};