Given a singly linked list of characters, write a function that returns true if the given list is a palindrome, else false.
METHOD 1 (Use a Stack):
- A simple solution is to use a stack of list nodes. This mainly involves three steps.
- Traverse the given list from head to tail and push every visited node to stack.
- Traverse the list again. For every visited node, pop a node from the stack and compare data of popped node with the currently visited node.
- If all nodes matched, then return true, else false.
Below image is a dry run of the above approach:
Below is the implementation of the above approach :
Python3
# Python3 program to check if linked# list is palindrome using stackclass Node: def __init__(self, data): self.data = data self.ptr = None # Function to check if the linked list# is palindrome or notdef ispalindrome(head): # Temp pointer slow = head # Declare a stack stack = [] ispalin = True # Push all elements of the list # to the stack while slow != None: stack.append(slow.data) # Move ahead slow = slow.ptr # Iterate in the list again and # check by popping from the stack while head != None: # Get the top most element i = stack.pop() # Check if data is not # same as popped element if head.data == i: ispalin = True else: ispalin = False break # Move ahead head = head.ptr return ispalin# Driver Code# Addition of linked listone = Node(1)two = Node(2)three = Node(3)four = Node(4)five = Node(3)six = Node(2)seven = Node(1)# Initialize the next pointer# of every current pointerone.ptr = twotwo.ptr = threethree.ptr = fourfour.ptr = fivefive.ptr = sixsix.ptr = sevenseven.ptr = None# Call function to check # palindrome or notresult = ispalindrome(one)print("isPalindrome:", result)# This code is contributed by Nishtha Goel |
Output:
isPalindrome: true
Time complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for using a stack, where n represents the length of the given linked list.
METHOD 2 (By reversing the list):
This method takes O(n) time and O(1) extra space.
1) Get the middle of the linked list.
2) Reverse the second half of the linked list.
3) Check if the first half and second half are identical.
4) Construct the original linked list by reversing the second half again and attaching it back to the first half
To divide the list into two halves, method 2 of this post is used.
When a number of nodes are even, the first and second half contain exactly half nodes. The challenging thing in this method is to handle the case when the number of nodes is odd. We don’t want the middle node as part of the lists as we are going to compare them for equality. For odd cases, we use a separate variable ‘midnode’.
Python3
# Python program to check if# linked list is palindrome# Node classclass Node: # Constructor to initialize # the node object def __init__(self, data): self.data = data self.next = Noneclass LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to check if given # linked list is palindrome or not def isPalindrome(self, head): slow_ptr = head fast_ptr = head prev_of_slow_ptr = head # To handle odd size list midnode = None # Initialize result res = True if (head != None and head.next != None): # Get the middle of the list. # Move slow_ptr by 1 and # fast_ptrr by 2, slow_ptr # will have the middle node while (fast_ptr != None and fast_ptr.next != None): # We need previous of the slow_ptr # for linked lists with odd # elements fast_ptr = fast_ptr.next.next prev_of_slow_ptr = slow_ptr slow_ptr = slow_ptr.next # fast_ptr would become NULL when # there are even elements in the # list and not NULL for odd elements. # We need to skip the middle node for # odd case and store it somewhere so # that we can restore the original list if (fast_ptr != None): midnode = slow_ptr slow_ptr = slow_ptr.next # Now reverse the second half # and compare it with the first half second_half = slow_ptr # NULL terminate first half prev_of_slow_ptr.next = None # Reverse the second half second_half = self.reverse(second_half) # Compare res = self.compareLists(head, second_half) # Construct the original list back # Reverse the second half again second_half = self.reverse(second_half) if (midnode != None): # If there was a mid node (odd size # case) which was not part of either # first half or second half. prev_of_slow_ptr.next = midnode midnode.next = second_half else: prev_of_slow_ptr.next = second_half return res # Function to reverse the linked list # Note that this function may change # the head def reverse(self, second_half): prev = None current = second_half next = None while current != None: next = current.next current.next = prev prev = current current = next second_half = prev return second_half # Function to check if two input # lists have same data def compareLists(self, head1, head2): temp1 = head1 temp2 = head2 while (temp1 and temp2): if (temp1.data == temp2.data): temp1 = temp1.next temp2 = temp2.next else: return 0 # Both are empty return 1 if (temp1 == None and temp2 == None): return 1 # Will reach here when one is NULL # and other is not return 0 # Function to insert a new node # at the beginning def push(self, new_data): # Allocate the Node & # Put in the data new_node = Node(new_data) # Link the old list of the new one new_node.next = self.head # Move the head to point to the # new Node self.head = new_node # A utility function to print # a given linked list def printList(self): temp = self.head while(temp): print(temp.data, end = "->") temp = temp.next print("NULL")# Driver codeif __name__ == '__main__': l = LinkedList() s = ['a', 'b', 'a', 'c', 'a', 'b', 'a'] for i in range(7): l.push(s[i]) l.printList() if (l.isPalindrome(l.head) != False): print("Is Palindrome") else: print("Not Palindrome") print()# This code is contributed by MuskanKalra1 |
Output:
a->NULL Is Palindrome b->a->NULL Not Palindrome a->b->a->NULL Is Palindrome c->a->b->a->NULL Not Palindrome a->c->a->b->a->NULL Not Palindrome b->a->c->a->b->a->NULL Not Palindrome a->b->a->c->a->b->a->NULL Is Palindrome
Time Complexity: O(n)
Auxiliary Space: O(1)
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