public static bool IsPalindrome(IntNode head)
{
if (head == null)
{
return(false);
}
// Traverse the list to count n
var nodeListLength = head.GetListLength();
if (nodeListLength == 1)
{
return(true);
}
// n == 2
// true iff elements are equal
//// Two node base case (not necessary)
//if (nodeListLength == 2)
// {
// if (head.Data != head.Next.Data)
// return false;
// }
// // Three node base case (not necessary)
// if (nodeListLength == 3)
// {
// if (head.Data != head.Next.Next.Data)
// return false;
// }
// Reverse first half of list 1/2 n
// if n < 4, remember don't need to put list back together at the end
// no reversing happens
// reverse 1/2 n nodes (left half of the nodes, not including the middle node if n is odd)
//if (nodeListLength > 1)
// {
// reverse numNodesToReverse nodes at beginning of list
// we'll undo this later to restore the list for the caller
var numNodesToReverse = nodeListLength / 2;
// This is true for both even and odd n
// I like Chris' idea of actually reversing the first half
// of the list in-place (rather than splitting it off as a
// seperate sublist, then needing to tack it back on again)
// Write logic to reverse k nodes each time. Then easy
// to use the same helper for both the original reverse,
// and the ultimate restore.
//Console.WriteLine(head);
//Console.WriteLine(numNodesToReverse);
//Console.WriteLine("\n");
// Returns head of half-reversed list
var leftRevHead = ReverseNodesAtStartOfNodeList(head, numNodesToReverse);
// Tail of reversed part of list is the head that was passed in
// Tail of the reversed part of the list is still hooked up to
// the non-reversed part of the list
//Console.WriteLine(leftRevHead);
var nodeAfterReversedPart = head.Next;
//var leftRevHead = prev;
// Handles finding proper start head for comparision (based on even or odd n)
IntNode rightHalfHead;
if (nodeListLength % 2 == 0)
{
// When n is even, start comparing at the two middlemost nodes
rightHalfHead = nodeAfterReversedPart;
}
else
{
// When n is odd, start comparing at the nodes to the left and right of
// the middlemost node
rightHalfHead = nodeAfterReversedPart.Next;
}
// Comparisons 1/2 n
var isPal = IntNodeListsDataAreEqual(leftRevHead, rightHalfHead);
// Undo the reverse 1/2 n
ReverseNodesAtStartOfNodeList(leftRevHead, numNodesToReverse);
// Not necessary anymore:
// Hook-up the right half of the original list
// to the left half of the (restored) original list
//leftRevHead.Next = nodeAfterReversedPart;
//Console.WriteLine($"leftRevHead == {leftRevHead}");
// Repoint what will be the restored tail now,
// calling the restore method
// It's the head of the (sub)list that is being resored
// Then the reverse method just does its job normally
// OR...do this after the helper finishes,
// as maybe the helper nulls the Next reference on that node
// (the undoPrev node in the helper, which is the leftRevHead outside the helper)
//}
return(isPal);
}
