Sort an array of 100 numbers so that there is an alternation of negative and positive elements of the array (C ++) If not difficult, then with comment lines.

  • And if they are not equally? That is, not 50 to 50? - AseN
  • there are 2 options. if you need a strict array sequence positive, negative, etc. then you must first sort the entire array. and then swap the queue 2 + 2*0 vs n - 1* (2*0) , 2 + 2 * 1 vs n - (1 + 2 * 1), 2 + 2 * 2 vs n - (1 + 2 * 2) ... `if you just need sorting with ignoring the value, then compare the absolute values ​​of the numbers (take sorting, for example, with a bubble on the wiki, and add elements of abs (value) when comparing the elements, - or whatever you have in C :)) - jmu
  • In this case, 50 to 50 - LickRosh

2 answers 2

In general, came up with 3 options. We work with the usual int array.

Option 1. The simplest, with the creation of a temporary array and back it up to the original array

 void put_mem (int* vec, size_t ss) { int temp [ss]; for (int i = 0, pi = 0, ni = 1; i < ss; i++) { if (vec[i] >=0) {temp[pi] = vec[i]; pi +=2;} else {temp [ni] = vec[i]; ni +=2;} } memcpy (vec, temp, sizeof(int)*ss); }

Option 2. With work directly in an array like bubble sorting. Introduced an additional helper function swapp for exchanging values ​​between array members. The algorithm is stupid and direct without optimization.

 static inline void swapp (int& a1, int& a2) { int temp = a1; a1 = a2; a2 = temp; } void put_sw (int* vec, size_t ss) { size_t j; for (size_t i = 0; i < ss; i++) { if (vec[i] < 0) { j = i; while (vec[++j] < 0); swapp (vec[i], vec[j]); } i++; if (vec[i] >= 0) { j = i; while (vec[++j] >= 0); swapp (vec[i], vec[j]); } } }

Option 3. The most sophisticated, with the use of STL. But it will work only with arrays whose size is not divisible by 4.

  void put_stl (int* vec, size_t ss) { int* it = partition (vec, vec+ss, bind2nd (less<int>(), 0)); for (int* itt = vec; it < vec+ss; it+=2, itt+=2) {swap (*itt, *it);} }

    In this code, if there are more negative or positive numbers, then they are simply added to the end, but what else? Do not discard them =)

     #include<iostream> using namespace std; void print_r(int *mass, int size) { for (unsigned i = 1; i <= size; i++) cout << mass[i] << ", "; cout << endl; } int main() { int mass[100]; for (unsigned i = 1; i <= 100; i++) mass[i] = rand() % 100 - 50; print_r(mass, 100); int h_m[100]; int s_m[100]; int c = 0; int c2 = 0; for (int i = 1; i <= 100; i++) { if (mass[i] < 0) { c2++; s_m[c2] = mass[i]; } if (mass[i] > 0) { c++; h_m[c] = mass[i]; } } cout << endl; print_r(h_m, c); cout << endl; print_r(s_m, c2); cout << endl; int new_mass[100]; int min = 0; if (c > c2) min = c2; else min = c; for (int i = 1; i <= min; i++) cout << s_m[i] << " " << h_m[i] << " "; if (min == c) for (int i = min + 1; i <= c2; i++) cout << s_m[i] << " "; if (min == c2) for (int i = min + 1; i <= c2; i++) cout << h_m[i] << " "; system("Pause"); }

    If the negative and positive elements are equally, then they can simply be displayed in a cycle from 1 to 50 alternately.