Max Points on a Line (6)

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

How can we determine a point on the same line? You can do this. Set an initial start point to calculate the slope of the straight line from the remaining point to the starting point. If the slope is the same, the points are on the same line. However, we also need to consider the absence of slope and repetition points. Therefore, the number and slope of repeat points are recorded separately as positive infinity. How can we obtain the final maximum value? It should be the number of points with the same slope, the number of repeated points and the number of lines in the vertical line, and the maximum number of repeated points, which should be clear for a while. To quickly obtain the number of straight lines with the same slope value, map is used in this example. Given the slope value, the corresponding correlation value can be quickly found, which is convenient and fast, you don't have to write a loop.

It was a bit embarrassing. After a long time, I finally finished it. Afterwards, I found that it was because I mistakenly wrote = in the if judgment condition. It was a bit sad! The ac will soon be available!

Paste the code now!

1 class Solution {2 public: 3 int maxPoints (vector <Point> & points) {4 if (points. size () <= 2) 5 return points. size (); 6 int I, j; 7 int coin; 8 int h; 9 int maxsize =-1; 10 double k; 11 for (I = 0; I <points. size ()-1; I ++) {12 map <double, int> point; 13 map <double, int>: iterator it; 14 h = 0; 15 coin = 1; // you have already done so. Because every line is formed, points start from 2! 16 for (j = I + 1; j <points. size (); j ++) {17 if (points [j]. x = points [I]. x & points [j]. y = points [I]. y) {18 coin ++; 19 continue; 20} 21 else if (points [j]. x = points [I]. x) {// This is where the = is converted to =! 22 h ++; 23 continue; 24} 25 else if (points [j]. y = points [I]. y) {26 k = 0.0; 27} 28 else {29 k = (double) points [j]. y-points [I]. y; 30 k/= (double) (points [j]. x-points [I]. x); 31} 32 it = point. find (k); 33 if (it = point. end () {34 point. insert (map <double, int >:: value_type (k, 1); 35} 36 else {37 point [k] ++; 38} 39 40} 41 h + = coin; // h must be greater than coin42 for (it = point. begin (); it! = Point. end (); it ++) {43 int temp = it-> second + coin; 44 if (temp> maxsize) 45 maxsize = temp; 46 47} 48 if (h> maxsize) 49 maxsize = h; 50 51} 52 53 return maxsize; 54 55 56} 57 };

 

Max Points on a Line (6)