Selection Sort : Java Program
Selection Sort is one of sorting techniques to get arranged
the numbers in ascending order.
Selection Sort method:
We will take the following list of elements.
10 30 15
6 4 1 8
Logic for finding the biggest number from the list:
Here we have 7 elements to be sorted. In the selection sort method, we will pick up
biggest element from the list. For this
we assume that first element is biggest number. Then compare the first number
with the second number in the list. Here
10 is the first number and 30 is the second number. From these two 30 is the
biggest number. Now we can treat 30 is the biggest number. Go for comparison
with the biggest number and next number 15. Again we get 30 is the biggest
number. Now compare biggest number 30 with the next number 6. Among these two
30 is the biggest number. Compare this biggest number with the next number 4. Now
compare 30 with the next number 1. Again we get 30 is the bigger number. Now
compare 30 with the last number 8. We get the 30 is the largest number from the
list.
The whole paragraph above gives that, finding of largest
number from the given list.
In the above list 30 is the biggest number and it is located
at 2 position. After finding the biggest number from the list, we will change the positions of biggest
number with the last number. That is we have to change positions of 30 and 8.
After exchanging the positions, the list
look like this
10 8 15
6 4 1 30
-------> I Pass
This is called first pass. During the first pass we can get
one element as sorted. Apply the same
logic to the remaining 6 elements.
Among the 6 elements 15 is the biggest element. Hence we have to swap the positions of
biggest number 15 and last number 1. After doing this, the list look like this.
10 8 1
6 4 15 30
--------> II Pass
This is called second pass.
Apply the same logic for the remaining 5 elements. The III
Pass looks like this after swapping 10 and 4 positions
4 8 1
6 10 15 30 --------> III Pass
Among the remaining four elements, 8 is the biggest number
and 6 is the last number. Therefore, we have to exchange these positions we get
the following the IV pass.
4 6 1 8
10 15 30 ---------à IV Pass
We have here 3 elements remaining. Among these 6 is the
largest one and last number is 1. We can swap these elements to get the V Pass
and it looks like the following
4 1 6 8
10 15 30 ---------à V Pass
We have here 4 is the largest element and 1 is the last
element. So we have swap these two to get the VI Pass. The list looks like
this.
1 4
6 8 10
15 30 --------------à
VI Pass
You can observe from the VI Pass, all the elements got
arranged in the ascending order.
If elements are 7, then 7-1=6 passes require to sort the
numbers in ascending order. In general,
if n elements are in the list, then n-1 passes are required to get the numbers
in sorted order.
This type of sorting technique is called selection sort
method.
Java Program for Selection Sort:
class Sele
{
public static void main(String args[])
{
int a[]={10,30,15,6,4,1,8};
int i,b,id,j,n;
n=a.length;
System.out.println(" Elements before sorting");
for(i=0;i<n;i++)
System.out.print(a[i] + " ");
for(i=n-1;i>=0;i--)
{
b=a[0];
id=0;
for(j=1;j<i+1;j++)
{
if(a[j]>b)
{
b=a[j];
id=j;
}
}
a[id]=a[i];
a[i]=b;
}
System.out.println();
System.out.println("Elements after sorting");
for(i=0;i<n;i++)
System.out.print(a[i] + " ");
}
}
output:
Elements before
sorting
10 30 15
6 4 1 8
Elements after sorting
1 4 6
8 10 15 30
