Array C++ Selection Sort

In this method, the smallest item in a given series of n elements is found and is placed in the first position. Then the smallest item from the remaining (n-l) elements is found and is placed in the second position and so on. This method requires (n-l) passes for arranging n elements.

Example

The program illustrates sorting numbers in ascending order using Selection sort.

#include <iostream.h>
const int m = 4;
void main() { 
  int data[m],i,j,k,min,minindex,n;
  cout<<"Enter the number of values "<<endl;
  cin>>n;
  cout<<"Enter the values "<<endl;
  for(i=0;i<n;i++)
    cin>>data[i] ;
    cout<<endl;
    cout<<"The list to be sorted is "<<endl;
    for(i=0;i<n;i++)
    cout<<data[i]<<" ";
    cout<<"\n" ;
    cout<<"Selection sort "<<endl;
    cout<<" ---------------" ;
    for(i=0;i<n;i++) {
      minindex=i;
      min=data[i] ;
      for(j=i+1;j<n;j++) {
        if(min>data[j]) {
           minindex=j;
           min=data[j] ;
         }
}
   data [minindex] =data [i];
   data[i]=min;
   cout<<endl<<"pass "<<i+1;
   for(k=0;k<n;k++ )
      cout<<" "<<data[k];
      cout<<endl;
}}

In the outer for loop, the variables min and minindex are initialized to i and data[i].In the first pass the minimum of all the elements, min is found. Index corresponding to minimum value is stored in minindex. Values corresponding to minindex and i are swapped, where i denotes the pass.

Input and Output:

Enter the number of values 5

Enter the values

67        22        58        49        10

The list to be sorted is

67        22        58        49       10

Selection sort

pass 1 10         22        58        49        67

pass 2 10         22        58        49        67

pass 3 10         22        49        58        67

pass 4 10         22        49        58        67

pass 5 10         22        49        58        67