No.149 Max Point on a Line

 1 /**
 2  * Definition for a point.
 3  * struct Point {
 4  *     int x;
 5  *     int y;
 6  *     Point() : x(0), y(0) {}
 7  *     Point(int a, int b) : x(a), y(b) {}
 8  * };
 9  */
10 class Solution 
11 {
12 public:
13     int maxPoints(vector<Point>& points)
14     {
15      //输入：二维平面上的一组点
16      //输出：同一条直线上的点的最大数量
17      //思路：两层循环，依次找出每两个点组成的直线的斜率slope，O(n2)本以为很大，还好，相信自己
18      //         将得到的数据存到map中：key为斜率，value为该斜率的点的个数
19         int count = points.size() ;
20         if( count == 0)
21             return 0;
22         if( count == 1)
23             return 1;
24         if( count == 2)
25             return 2;
26         
27         map<double, int> result;
28         double slope;
29         for(int i=0; i< count ; i++)
30             for(int j = i+1; j<count; j++)
31             {
32                 if(points[i].x == points[j].x)
33                     slope = numeric_limits<double>::max();
34                 else
35                     slope = (points[i]. y - points[j]. y)/(points[i]. x - points[j]. x);//除法!!分母
36                 if(result.find(slope) == result.end())
37                     result[slope] = 2;
38                 else
39                     result[slope]++;
40             }
41         
42         int max = 0;
43         for(auto const i : result)
44         {
45             if(i.second > max)
46                 max = i.second;
47         }
48         return max;
49     }
50 
51 };