CS307 Fall 2004 Midterm 2 Suggested Solutions

1. Short answer questions. Answer as below or -2. On Big O okay if leave off the O()

A.	12

B.	*1 2 3 4

C.	45

D.	O(N)

E.	O(log N)

F.	O(1)

G.	O(N^2)

H.	O(N + M) //O(N) or O(M) okay as well

I.	O(N^2)

J.	15 sec

K.	24 sec

L.	104 sec. (fractions okay, no terms with log allowed)

M.	5 6 7 8 9

N.	13 2 5 11

O.	9 4 4 1 8 0

2.
Set result = new Set();
for(int i = 0; i < myCon.size(); i++)
	if( !otherSet.myCon.contains( myCon.get(i) )
		result.myCon.add( myCon.get(i) );
return result;

3.
boolean result;
// other list empty? if so this contains all
if( other.myHead == null )
	result = true;
else
{	SortableNode thisTemp = myHead;
	SortableNode otherTemp = other.myHead;
	int res;
	while( thisTemp != null && otherTemp != null )
	{	res = otherTemp.getData().compareTo( thisTemp.getData() );
		//current data in other larger than data in this
		if( res > 0 )
			thisTemp = thisTemp.getNext();
		//if == advance otherTemp (allow double counting if necessary)
		//if ensuring one for one I would advance both pointers
		else if( res == 0 )
			otherTemp = otherTemp.getNext();
		// current data in other smaller (!) than data in this
		// because sorted it will not be present and we can stop
		else
			thisTemp = null;
	}
	//had to "fall off" other to have counted them all.
	result = otherTemp = null;
}
return result;

4.	
return distHelper(myHead, 0);

private int distHelper(Node cur, int distSoFar)
{	int result;	
	//found it?	
	if( cur == myTail )
		result = distSoFar;
	//fell off? if so send back flag to indicate dead end
	else if( curr == null )
		result = -1;
	else
	{	//check next and add 1 for moving forward	
		result = distHelper( cur.getNext(), distSoFar + 1);
		// Did that lead to a dead end? 
		// If so check the data node and if it is a Node, try it
		if( result == -1 && cur.getData() instanceof Node)
			result = distHelper( (Node)cur.getData(), distSoFar + 1);
	}
	return result;
}