Leetcode 149 Max Points on a Line

对于点来说，分三种情况：

1.普通点：具有斜率。

2.垂直于X轴的点：斜率无穷大。

3.坐标相同的点。

 

我们将1和2分别进行讨论，然后和3的个数分别相加，求出较大者，即为对于一个点来说，和它共线的最多的点的个数。

再通过循环得出最大值即可。

 

class Solution {
public:
    double getGradient(Point p1, Point p2) {
        return (double)(p2.y - p1.y) / (p2.x - p1.x);
    }
    
    
    int maxPoints(vector<Point>& points) {
        int n = points.size();
        
        if (n <= 1) {
            return n;
        }
        
        int res = 0;
        
        for (int i=0; i<points.size(); ++i) {
            int localmax = 1;
            int samePoints = 0;
            int sameX = 0;
            int sameGradient = 1;
            unordered_map<double, int> hashmap;
            
            for(int j=0; j<points.size() && i != j; ++j){
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    ++samePoints;
                    continue;                
                }
                
                if (points[i].x == points[j].x) {
                    if (sameX == 0) {
                        sameX = 2;
                    }
                    
                    else {
                        ++sameX;
                    }
                }
                
                else {
                    double gradient = getGradient(points[i], points[j]);
                    if (hashmap[gradient] == 0) {
                        hashmap[gradient] = 2;
                    }
                    
                    else {
                        ++hashmap[gradient];
                    }
                }
            }
            
            for(unordered_map<double, int>::iterator it = hashmap.begin(); it!=hashmap.end(); ++it) {
                sameGradient = max(sameGradient, it->second);
            }
            
            localmax = max(sameGradient + samePoints, sameX + samePoints);
            res = max(res, localmax);
        }
        
        return res;
    }
};