[skyline2162]

I do not understand the selection sort code and how it works to put numbers in acceding order. I do not understand why its array.length-1 for the size, doesn't it need to search the last array element as well? I understand the process how the lowest number is switched eg. 1, 5 ,7, 2 would become 1, 2, 7, 5 teb 1 , 2, 5, 7, but I just do not understand the code behind it.

public static void selectionSort(int[] array)
{
	int startScan;
	int index;
	int minIndex;
	int minvalue;
	
	for (startScan = 0; startScan < (array.length-1); startScan++)
	{
		minIndex = startScan;
		minValue = array[startScan];
		
		for (index = startScan + 1; index < array.length; index++)
		{
			if (array[index] < minValue)
			{
				minValue = array[index]
				minindex = index;
			}
		}
	
	
	array[minIndex] = array[startScan];
	array[startScan] = minValue;
	}
}

The only way a binary search would work is if it is sorted in acending order so that would be the slection sort, so that would go before the binary search? Also for the binary search why is the last number array.length-1 and not just array.length.

public static int binarySearch(int[] array, int value)
{
	int first;
	int last;
	int middle;
	int position;
	boolean found;
	
	first = 0;
	last = array.length -1;
	position = -1;
	found = false;
	
	while (!found && first <= last)
	{
		middle = (first + last ) / 2;
		
		if (array[middle] == value)
		{
			found = true;
			position = middle;
		}
		else if (array[middle] > value)
		{
			last = middle -1;
		}
		else
		{
			first = middle +1;
		}
	}
	
	return position;
}

Thank you

[macosxnerd101]

The way the selection sort algorithm works is by choosing the smallest number as the first element in the array. Then it evaluates the rest of the elements, updating the smallest number as necessary. The smallest number is then swapped with the first element in the list, completing one iteration of the outer loop. Since element 0 is in place, we only need to evaluate elements 1 through length-1.

Quote

The only way a binary search would work is if it is sorted in acending order so that would be the slection sort, so that would go before the binary search?

The sort would need to occur before the search.

Quote

Also for the binary search why is the last number array.length-1 and not just array.length.

Arrays are indexed from through length-1. So if you have 10 elements, they are indexed from 0-9. If you access index 10, you will go out of bounds.

[skyline2162]

but doesn't the array.length always incorporate from 0 to that number?

[macosxnerd101]

No. If you have ten elements, that's the length. Those elements are indexed 0-9.

[skyline2162]

thank you!

[macosxnerd101]

Glad I could help!