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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 | #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
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