Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
思路:穷举法,双层循环,用map实现
不过需要注意的一点,不要忘记点是重复的点,以及考虑斜率无穷大的情况
代码:
/**
* Definition for a point.
* 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) {
if(points.empty())
return ;
unordered_map<float,int> mp;
int max_num = ;
for(int i = ;i < points.size();i++){
mp.clear();
mp[INT_MAX] = ;
int duplicated = ;
for(int j = i+;j < points.size();j++){
if(points[i].x==points[j].x && points[i].y==points[j].y){//重复点
duplicated++;
continue;
}
//如果没有进上面的if,说明不是重复点,x又相等的话,斜率无穷大
float k = points[i].x==points[j].x ? INT_MAX : (float)(points[j].y - points[i].y)/(points
[j].x - points[i].x);
mp[k]++;
}
for(unordered_map<float,int>::iterator mit = mp.begin();mit!=mp.end();mit++){
if(mit->second + duplicated > max_num)//需要重复点和计数加起来作为这条斜率上的点
max_num = mit->second + duplicated;
}
}
return max_num;
}
};