c++ - Recursive Way to Find 2nd/3rd Smallest Element In Array -

below code written find smallest element of unsorted array in c++.

how should go editing recursive function find second or third smallest element in array?

i understand there other methods out there (i've searched), i'd know how using recursive function works minimum number.

int min(vector<int> &array, int &min, int &next_min, int left,int right) {         int a;             int b;      if(left == right) {          return array[left];     } else {            int mid= (right+left)/2;          = min(array,left,mid);         b = min(array,mid+1,right);          if (a<b)              return b;         else                      return a;     } } 

many in advance

here attempt @ finding 2nd smallest:

#include<iostream> #include<vector> #include <cstdlib>  using namespace std;  int counter=0;  void minf(vector<int> &array,int left,int right,int &min, int &next_min);  int main() {     int next_min;      cout << "enter integers 1 line @ time. (enter -999 terminate)" << endl;      vector<int> array;  //      int input;      while(true){         cin >> input;         if(input == -999) //check termination condition         break;         array.push_back(input);     }      int imin;     int imax;      cout<<"enter imin, imax"<<endl;     cin>>imin;     cin>>imax;       cout<<"min number " << next_min <<endl;      cin.get();     system("pause");     return 0; }  void minf(vector<int> &array,int left,int right,int &min, int &next_min) {     int a;     int b;     int c;     int d;      if(left == right) {          min = array[left];         next_min = 2147483647;     } else {         int mid= (right+left)/2;          minf(array,left,mid,a,b);         minf(array,mid+1,right,c,d);          if (a < b && < c && < d) {             min = a;             if (b<c && b <d)                 next_min = b;             else if (c < b && c < d)                 next_min = c;             else                 next_min = d;         }              if (b < && b < c && b < d){             min = b;             if (a<c && <d)                 next_min = a;             else if (c < b && c < d)                 next_min = c;             else                 next_min = d;         }          if (c < && c < b && c < d) {             min = c;             if (b<a && b<d)                 next_min = b;             else if (a < b && < d)                 next_min = a;             else                 next_min = d;         }             if (d < && d < c && d < b){             min = d;             if (a<c && <b)                 next_min = a;             else if (c < b && c < a)                 next_min = c;             else                 next_min = b;         }          } } 

this function find n smallest elements in recursive way, hope helps!

#include <vector> #include <iostream>  using namespace std;  vector<int> nmin(const vector<int> &array, int n, int left, int right) {     vector<int> result;     if (left == right) {         result.push_back(array[left]);     } else {         int mid = (right + left) / 2;         vector<int> leftresult = nmin(array, n, left, mid);         vector<int> rightresult = nmin(array, n, mid + 1, right);         int = 0;         int l = 0;         int r = 0;         int l = leftresult.size();         int r = rightresult.size();         while (i < n && (l < l || r < r)) {             i++;             if (l < l) {                 if (r < r) {                     if (leftresult[l] < rightresult[r]) {                         result.push_back(leftresult[l++]);                     } else {                         result.push_back(rightresult[r++]);                     }                 } else {                     result.push_back(leftresult[l++]);                 }             } else {                 result.push_back(rightresult[r++]);             }         }     }     return result; }  int main() {     vector<int> test = {-2, 6, 7, 1, 3, 7, 4, 2, 5, 0, 8, -2};     vector<int> smallest3 = nmin(test, 3, 0, test.size() - 1);     (int num : smallest3) {         cout << num << endl;     } } 

brief explanation: n smallest elements in ascending order left & right half, call them leftresult , rightresult, merge two, picking next smallest element 2 partial results. merge sort, except return n elements instead of sorting whole array.