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1. ## Palindrome using stack,queue

Need some help with a little coding problem.
I need to check if it is a Palindrome using a queue and a stack, by implementing a method that tests whether a sequence of numbers is a palindrome, that is, equal to the sequence in reverse.
the code in [...] is My Code, while everything else is unchangeable any help is appreciated.
Code:
import java.util.Queue;
import java.util.Stack;

public class QueueStackUtil
{
public static boolean isPalindrome(int[] values)
{
Stack<Integer> stack = new Stack<Integer>();
for (int s : values)
{
// insert s into the queue and stack
[//*MyCode//ins instack
s[++queue]=values;
//ins in queue
s[++stack]=values;]
}
while (queue.size() > 0)
{
[MyCode// remove an element from the queue
if(!isPaindrome()){
for(int i=0; i<tail;i++)
queue[i] = queue[i+1];
tail--;
}
// remove an element from the stack
if(!isPalindrome()) stack--;

// compare the removed elements
if(stack==queue)
!isPalindrome]

}
return true;
}
}

• Making use of both a queue and a stack, this is the shortest I could come up with:
PHP Code:
import java.util.Deque;

public class
Palindrome
{
private
private
private
String s;
public
Palindrome(String s)
{

this.s;
for (
Character c s.toCharArray())
{

Character.toLowerCase(c);
if (
Character.isLetterOrDigit(c))
{

}
}
}

public
boolean isPalindrome()
{
return
this.stack.equals(this.queue);
}

public
String getString()
{
return
this.s;
}

public static
void main(String[] argv)
{

System.out.println("'" p.getString() + "' is" + (p.isPalindrome() ? "" " not") + " a palendrome");
}

Realistically, if I were to check for a palindrome I would would grab a string from the middle and compare outside moving both backwards and forwards, and skip over any non letter or digit. That would let me drop the iterative count to 1/2 that of the string length if it is a palindrome.

Since your code is clearly homework, you'll need to do some adaptations here. First, you're stuck using the queue and stack classes (which are not recommended). LinkedList is easy since it implements the Deque, so you can actually treat it as both a stack and a queue.
This makes no sense: s[++queue]=values. Values represents the entire int[], and s[x] represents an invalid offset to a numerical data. This syntax: for (int s : values) is java's equivalent to a foreach. So the 's' here represents a single integer value which you need to push onto the lists. To add items to your stack, use the stack.push method. To add to the queue, use the queue.offer or queue.offerLast method.

Now in the while loop, you simply capture the results of queue.poll and stack.pop. If they don't match, you need to return false.
That's a little curious I have to admit, more often than not educational facilities prefer a single point of return and opt for a variable instead of an inline return.

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