leetcode Max Points on a Line

在一个平面内有很多点，返回在同一条直线上的最多点数。

用map来记录通过某点的，和其他点构成的斜率有几个点。因为斜率有不存在的情况，另外用一个数记录。还有，因为可能会有重复的点，所以重复的点也是要另外记录的。最后返回最大值。

用迭代器遍历map，找出最大的数。因为map可能是零，因为如果所有的点够成的斜率为无穷大，那么就可能木有map了，所以斜率无穷大作为另外记录的要注意。

还有是map可以通过first和second来访问key和value

/**
* 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) {
unordered_map<double, int> umap;
int size = points.size();
if (size == 0 || size == 1) return size;
int samePoint = 1, inSlope = 1, ans = -1; // infinite slope

for (int i = 0; i < size - 1; i++)
{
for (int j = i + 1; j < size; j++)
{
if (points[i].x == points[j].x)
{
if (points[i].y == points[j].y)
samePoint++;
inSlope++;
continue;
}
double slope = (1.0 * (points[i].y - points[j].y) / (points[i].x - points[j].x));
umap[slope]++;
}
// find the maxmum
for (unordered_map<double, int>::iterator itr = umap.begin(); itr != umap.end(); ++itr)
{
if (itr -> second + samePoint > ans)
ans = itr -> second + samePoint;
}
if (inSlope > ans)
ans = inSlope;
umap.clear();
inSlope = 1;
samePoint = 1;
}
return ans;
}
};