Max points on a line

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

For two points (x1, Y1) and (X2, Y2), the linear equation is: (x1-x2) * Y + (y2-y1) * x + x2 * y1-x1 * y2 = 0, that is: A * Y + B * x + c = 0.

Using the idea similar to breadth-first traversal, Q [I] stores the straight line between the I + 1 vertex and each vertex in the first I vertex, and the number of vertices passing through the straight line in the first I + 1 vertex, Q [I] [0] indicates the number of times that vertex I + 1 has already appeared.

Processing each element of the input sequence in turn, for the current element I, sequentially accessing each Q [k] (k = I-1, I-2 ,..., 0), traverse Q [k] To find the maximum number of vertices in the straight line determined by point I and point K, and save them in Q [I.

In the process of accessing Q [k] from the back, the next value of Q [I] [0] is updated when I and I are duplicated for the first time.

 1 class Line 2 { 3  public: 4     Line(): a(0), b(0), c(0), cnt(1) {} 5     ~Line(){} 6  public: 7     int a, b, c; 8     int cnt; 9 };10 11 class Solution {12 public:13     int maxPoints( vector<Point> &points ) {14         if( points.size() <= 2 ) { return points.size(); }15         vector<Line>* Q = new vector<Line>[points.size()];16         Q[0].resize(1);17         int max_num = 2;18         Line optimal;19         for( int i = 1; i < points.size(); ++i ) {20             Q[i].resize(1);21             for( int j = i-1; j >= 0; --j ) {22                 optimal.a = points[i].x-points[j].x;23                 optimal.b = points[j].y-points[i].y;24                 optimal.c = points[j].x*points[i].y-points[i].x*points[j].y;25                 if( points[i].x == points[j].x && points[i].y == points[j].y ) {26                     if( Q[i][0].cnt == 1 ) { Q[i][0].cnt = Q[j][0].cnt+1; }27                     else { continue; }28                     optimal.cnt = Q[i][0].cnt;29                     for( size_t ID = 1; ID != Q[j].size(); ++ID ) {30                         Line& aLine = Q[j][ID];31                         if( aLine.a*points[i].y + aLine.b*points[i].x + aLine.c == 0 ) {32                             if( optimal.cnt <= aLine.cnt+1 ) { optimal = aLine; ++optimal.cnt; }33                         }34                     }35                 } else {36                     optimal.cnt = 2;37                     for( size_t ID = 0; ID != Q[j].size(); ++ID ) {38                         Line& aLine = Q[j][ID];39                         if( aLine.a*points[i].y + aLine.b*points[i].x + aLine.c == 0 ) {40                             if( optimal.cnt < aLine.cnt+1 ) { optimal.cnt = aLine.cnt+1; }41                         }42                     }43                 }44                 if( max_num < optimal.cnt ) { max_num = optimal.cnt; }45                 Q[i].push_back( optimal );46             }47         }48         delete[] Q;49         return max_num;50     }51 };