Max Points on a Line

Max Points on a Line

• 题目链接：http://oj.leetcode.com/problems/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.

• 题目大意：给定n个二维平面上的点，求出这n个点当中最多有多少个点在同一直线上。
• 解题思路：每次取定一个点p1，然后求出p1与其他的n-1个点的斜率，并统计斜率相同的数目，取最大值即可。
• 具体实现：

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

#include <iostream> #include <map> #include <climits>   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 maxPoints(vector<Point> &points) {         int maxNum =0;         int size = points.size();         int i,j;         if(size <=2)   //如果点的个数小于3个，则最大数目为点的个数             return size;         for(i=0;i<size;i++)         {             int cnt =0;             map<double,int> mp;             for(j=0;j<size;j++)             {                 if(i==j)                     continue;                 if(points[i].x == points[j].x && points[i].y == points[j].y)                 {                     cnt++;                     continue;                 }                   //注意当直线与y轴平行的情况                 double slope = points[i].x == points[j].x ? INT_MAX : (double)(points[j].y - points[i].y)/(points[j].x - points[i].x);                  mp[slope]++;              }                           if(mp.size()==0)   //防止mp为空时，下面的for循环不执行                 mp[INT_MIN] =0;             map<double,int>::iterator it = mp.begin();             for(;it!=mp.end();it++)             {                 if(it->second+ cnt > maxNum)                     maxNum = it->second +cnt;             }         }         return maxNum+1;     } };     int main(int argc, char *argv[]) {     vector<Point> points;     Point point[3];     int i=0;     for(i=0;i<3;i++)     {         point[i].x = 1;         point[i].y= 1;         points.push_back(point[i]);     }     Solution solution;     cout<<solution.maxPoints(points);     return 0; }

　　这里要注意3个地方：

　　1）如果点的个数小于3个，则最大数目为点的个数；

　　2）考虑重复点的情况，重复点是无法计算斜率的；

　　3）考虑直线与y轴平行时，斜率为无穷大的情况。