力扣算法题—149Max Points on a line

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Example 1:

Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

Example 2:

Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Solution：

　　直接求每个点与其他点的斜率，然后计算相同斜率的个数就行

　　注意两点：

　　一个是会有相同的点出现，另一个是斜率计算得到无穷和double比较相等的情况

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        if (points.size() < 3)return points.size();
        int res = 0;        
        for (int i = 0; i < points.size(); ++i)
        {
            int same = 1;
            for (int j = i + 1; j < points.size(); ++j)
            {
                int cnt = 0;
                long long int x1 = points[i][0], x2 = points[j][0];
                long long int y1 = points[i][1], y2 = points[j][1];
                if (x1 == x2 && y1 == y2) { ++same; continue; }
                for (int k = 0; k < points.size(); ++k)
                {
                    long long int x3 = points[k][0], y3 = points[k][1];
                    if ((x1*y2 + x2 * y3 + x3 * y1 - x3 * y2 - x2 * y1 - x1 * y3) == 0)
                        ++cnt;
                }
                res = max(res, cnt);
            }
            res = max(res, same);
        }
        return res;
    }
};