Max Points on a Line(转载,完全转载hackersun)
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/** * 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) { const int n = points.size(); unordered_map<float,int> m; int maxRes=0; for(int i=0;i<n;++i) { m.clear(); m[INT_MIN]=0; int duplicates=1; for(int j=0;j<n;++j) { if(i==j) continue; if(points[i].x==points[j].x && points[i].y==points[j].y) { duplicates++; continue; } float k=points[i].x==points[j].x ? INT_MAX: (float)(points[j].y-points[i].y)/(points[j].x-points[i].x); m[k]++; } unordered_map<float, int>::iterator iter = m.begin(); for(;iter!=m.end();++iter) { if(iter->second+duplicates>maxRes) maxRes=iter->second+duplicates; } } return maxRes; }};注意:
0、points中重复出现的点。
1、int maxNum = 0;
初始化,以防points.size() ==0的情况。
2、mp[INT_MIN] = 0;
保证poins中只有一个结点,还有points中只有重复元素时,mp中没有元素。这两种极端情况。
3、int duplicate = 1;
duplicate记录重复点的数量,初始化为1,是因为要把当前的点points[i]加进去。
4、float k = points[i].x == points[j].x ? INT_MAX : (float)(points[j].y - points[i].y)/(points[j].x - points[i].x);
计算斜率,如果直线和y轴平行,就取INT_MAX,否则就取(float)(points[j].y - points[i].y)/(points[j].x - points[i].x)
一开始把(float)(points[j].y - points[i].y)/(points[j].x - points[i].x)写做(float)((points[j].y - points[i].y)/(points[j].x - points[i].x))一直就不对,后来才想明白,注意注意!
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