How to swaps first and last element of a queue in C++? - c++

Question

I have written a function to swap first and last elements of a queue in C++
void swap(queue Q)
{
queue temp;
createQ (temp);
int x,first,last;
first=dequeue(Q);
while(!isemptyQ(Q))
{
x=dequeue(Q);
last=x;
enqueue(x,temp);
}
enqueue(last,Q);
while(!isemptyQ(temp))
{
x=dequeue(temp);
if(x!=last) enqueue(x,Q); //(if) to avoid adding last element again
}
enqueue(first,Q)
}
note: "just consider (isempty,dequeue,createQ and enqueue) are other functions>>(my first year)"
but if the last element was duplicated in the middle like that
if that are the elements of the queue (5,1,9,3,9)
if we trace that function on these elements it would be (9,1,3,5)
there's a 9 missing !!
so if there's an escape from that or any other idea for the function ?!

Answer 1

No need to check for last if you don't push it.o I would rather do something like:
while(!isemptyQ(Q))
{
x=dequeue(Q);
if (isemptyQ(Q))
{
last=x;
}
else
{
enqueue(x,temp);
}
}
enqueue(last,Q);
while(!isemptyQ(temp))
{
x=dequeue(temp);
enqueue(x,Q);
}
By the way, your code won't work for empty or 1 element queues.

Answer 2

Below is pseudo code that tries to cover all special cases including 1 or zero elements in the queue
swap_first_last(q) {
dequeue(q, x, isEmpty);
initialize(tmp);
first = x;
count = 0;
while(!isEmpty) {
count ++;
last = x;
dequeue(q, x, isEmpty);
if(!isEmpty) {
enqueue(tmp, last);
}
}
if(count == 0) return;
enqueue(q, last);
if(count >= 2) {
dequeue(tmp, x, isEmpty);
while(!isEmpty) {
dequeue(tmp, x, isEmpty);
if(!isEmpty) {
enqueue(q, x);
}
}
enqueue(q, first);
}
}