leetcode面试题3：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.

就是算在同一条直线上的点有多少个，需要注意的有几点：

1.坐标相同的点算在同一条直线上

2.斜率相同的点在同一条直线上，但特例是横坐标相同的点斜率不存在

思路是对每个点遍历剩余的点，创建一张哈希表map<double,int>，存储每个斜率对应的直线上的点数，斜率不存在时对应key值INT_MAX

AC代码：

class Solution {public:    int maxPoints(vector<Point> &points) {        double k;        int i,j,max=0;        if(points.size()<=2)return points.size();        for(i=0;i<points.size();i++)        {            map<double,int>count;            count[INT_MIN]=0;            int samepoint=1;            for(j=0;j<points.size()&&j!=i;j++)            {                if(points[j].x==points[i].x&&points[j].y==points[i].y)                samepoint++;                else                {                    double k=points[j].x==points[i].x?INT_MAX:(double)(points[j].y-points[i].y)/(double)(points[j].x-points[i].x);                    count[k]++;                }            }            map<double,int>::iterator ite = count.begin();            for(ite;ite!=count.end();ite++)            {                if(ite->second+samepoint>max)                max=ite->second+samepoint;            }        }        return max;    }};