1 class Solution {
 2 public:
 3     int maxPoints(vector<Point>& points) {
 4         map<double,int> slope;           //建立斜率与出现次数的映射
 5         if(points.size() <= 2)          //点的个数小于3时，直接返回点的个数
 6             return points.size();
 7          int n = points.size(),ans = 0;     //ans为最终结果
 8         for(int i = 0; i < n - 1; i++)
 9         {
10             int maxNum = 0;                //记录当前节点对应的斜率出现最多的次数
11             int same_x = 1;                //记录斜率不存在，即横坐标相等的情况
12             int samePoint = 0;             //记录重合的点
13             slope.clear();
14             for(int j = i + 1; j < n; j++)
15             {
16
17                   if (i == j)
18                       continue;
19                   if (points[i].x == points[j].x && points[i].y == points[j].y)
20                   {
21                     samePoint++;
22                     continue;
23                 }
24                   if (points[i].x == points[j].x)  //斜率不存在
25                   {
26                       same_x++;
27                       continue;
28                   }
29                   double k = (double)(points[i].y-points[j].y)/(points[i].x-points[j].x);
30                   if (slope.find(k) != slope.end())   //更新当前斜率出现的次数
31                       slope[k]++;
32                   else
33                       slope[k] = 2;
34             }
35             for (map<double,int>::iterator it = slope.begin(); it != slope.end(); ++it)   //找出最大次数的斜率
36                 same_x = max(same_x,it->second);
37             maxNum = same_x + samePoint;
38             ans = max(ans,maxNum);                  //与之前的最大值比较
39         }
40         return ans;
41     }
42 };