149 Max Points on a Line

链接：https://leetcode.com/problems/max-points-on-a-line/
问题描述：
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

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这个问题就是求给出一些点的坐标，求在同一条直线上最多有几个点。直线表达式为y=kx+b，简单的解法就是遍历，取一个点p1，求出和其他点的斜率，如果斜率相同，那么这些点一定在同一直线上。需要注意的是斜率可能不存在情况。还有可能有和p1相同的点。自己用vector记录k和同线的点的数目不仅写代码比较麻烦，还不容易理解。

 struct data {     double k;     int numofpoint; };class Solution {public:    int maxPoints(vector<Point>& points) {        if(points.size()<3)            return points.size();        int result=0,numofsamepoint=0;        vector<data> m;        data temp;        for(int i=0;i<points.size();i++)        {            for(int j=i+1;j<points.size();j++)            {                if(points[i].x==points[j].x&&points[i].y==points[j].y)                {                       numofsamepoint++;                    continue;                }                else if(points[i].x==points[j].x)                {                    temp.k=INT_MAX;                    temp.numofpoint=2;                    if(m.size()==0)                     m.push_back(temp);                    else                    for(int k=0;k<m.size();k++)                    {                        if(m[k].k==INT_MAX)                        {                         m[k].numofpoint++;                         break;                        }                        if(k==m.size()-1)                       {                            m.push_back(temp);                            break;                       }                    }                }                else                {                   temp.k=double(points[i].y-points[j].y)/double(points[i].x-points[j].x);                     temp.numofpoint=2;                   if(m.size()==0)                       m.push_back(temp);                   else                   for(int k=0;k<m.size();k++)                   {                       if(m[k].k==temp.k)                       {                           m[k].numofpoint++;                           break;                       }                       if(k==m.size()-1)                       {                            m.push_back(temp);                            break;                       }                   }                }               }            result= result>numofsamepoint+1?result:numofsamepoint+1;            for(int k=0;k<m.size();k++)                       if(m[k].numofpoint+numofsamepoint>result)                      result=m[k].numofpoint+numofsamepoint;                m.clear();                numofsamepoint=0;        }    return result;    }};

用C++的unordered_map可以更好的解决这个问题。

class Solution {public:    int maxPoints(vector<Point>& points) {        if(points.size()<3)            return points.size();        int result=0,numofsamepoint=1;        unordered_map<float,int> m;        for(int i=0;i<points.size();i++)        {            for(int j=i+1;j<points.size();j++)            {                if(points[i].x==points[j].x&&points[i].y==points[j].y)                {                       numofsamepoint++;                    continue;                }                else                    m[points[i].x == points[j].x ? INT_MAX : (float)(points[j].y -                    points[i].y)/(points[j].x - points[i].x)]++;            }            result= result>numofsamepoint?result:numofsamepoint;            unordered_map<float, int>::iterator it = m.begin();              for(; it != m.end(); it++)                  if(it->second + numofsamepoint > result)                        result=it->second + numofsamepoint;            m.clear();            numofsamepoint=1;        }    return result;    }};