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 program to check if linked # list is palindrome using stack class Node:
def __init__( self , data):
self .data = data
self .ptr = None
# Function to check if the linked list # is palindrome or not def 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 list one = 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 pointer one.ptr = two
two.ptr = three
three.ptr = four
four.ptr = five
five.ptr = six
six.ptr = seven
seven.ptr = None
# Call function to check # palindrome or not result = 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’.
# Python program to check if # linked list is palindrome # Node class class Node:
# Constructor to initialize
# the node object
def __init__( self , data):
self .data = data
self . next = None
class 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 code if __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)
Please refer complete article on Function to check if a singly linked list is palindrome for more details!