Vector removal

Got it.
int
The vector, the elements with the plain indexes of which contain the coordinates x of the nearest points, and the inaccurate coordinates from the same points. The coordinates are either 0 or 1, i.e. the point(0.3)(1.3) neighbours and (1.2),(3,2), no ( see example*).Example:
1,1,1,2,2,2,2,1
 Point vector(1,1),(1,2),(2,2),(2,1)
the coordinates x and y of the following points are different from those of 0 or 1; the figure obtained is a square; it could be either11122221
or12222111
 It's important that points go straight without repeating.Well, if we've got a vector, for example.
1,1,2,2,3,3,4,2,5,1,4,1,3,1,2,1
i.e. triangle on points (1,1),(2,2),(3,3),(4,2),(5,1),(4,1),(3,1),(2,1)
with summit(1,1),(3,3),(5,1)
I'd like to get a vector from the original vector with these three peaks, that is.1,1,3,3,5,1
, i.e. remove intermediate pointswhich "no role" from the vector (for any figure, not just a triangle).Here's my code to remove these points:
i=0; // текущий индекс while (i<v.size()1) // до последней координаты x { int prevX=(!i ? v[v.size()2] : v[i2]),nextX=(i==v.size()2 ? v[0] : v[i+2]),prevY=(!i ? v[v.size()1] : v[i1]),nextY=(i==v.size()2 ? v[1] : v[i+3]); // получаем предыдущие и следующие координаты if (v[i]==(prevX+nextX)/2 && v[i+1]==(prevY+nextY)/2){ // если условие удаления выполняется v.erase(v.begin()+i,v.begin()+i+2); // вот здесь не уверен, пробовал два удаления подряд  безрезультатно vector<int>(v).swap(v); // очистка памяти } else i+=2; // продвигаем индекс на следующую координату x (при удалении индекс сам "продвигается") }
The point is that the coordinates of the current point are checked (in the vector)
v[i]
(d) Mean arithmetic for the previousv[i2]
and followupv[i+2]
coordinates x. Same for y.He doesn't work, I suspect it's a mistake in the distance itself. Please indicate the error/and.

Remove the divide, write 2*v[i]==. ♪ ♪