Max Points on a line

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

思路：這道題主要是找出重疊結點和相同斜率之和。此時我們要判斷有斜率和無斜率個數，並判斷那個數量比較大，更新斜率最大值。然後在與相同情況下重疊結點個數累加。最後就是更新同一條直線上點數最大值。

/** * 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) {        if(points.size() <= 2){            return points.size();        }                map<double,int> slop_lines;        int maxNumber = 0, same_points, none_slop;        double slop;        for(int i = 0; i < points.size()-1; ++i){            slop_lines.clear();            same_points = 1;            none_slop = 0;            for(int j = i+1; j < points.size(); ++j){                if(points[i].x == points[j].x &&  points[i].y == points[j].y)                {                    ++same_points;                    continue;                }                if(points[i].x == points[j].x){                    ++none_slop;                }else{                    slop = (double)(points[i].y - points[j].y)/(points[i].x - points[j].x);                    if(slop_lines.find(slop) != slop_lines.end()){                        ++slop_lines[slop];                    }else{                        slop_lines[slop] = 1;                    }                }            }                        //更新最大值            map<double,int>::iterator iter = slop_lines.begin();            for(;iter != slop_lines.end(); ++iter){                if(iter->second > none_slop){                    none_slop = iter->second;                }            }            none_slop += same_points;            if(none_slop > maxNumber){                maxNumber=none_slop;            }        }                return maxNumber;    }};