C++ Program To Check If A Singly Linked List Is Palindrome
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 :
C++
#include<bits/stdc++.h>
using namespace std;
class Node
{
public :
int data;
Node( int d)
{
data = d;
}
Node *ptr;
};
bool isPalin(Node* head)
{
Node* slow= head;
stack < int > s;
while (slow != NULL)
{
s.push(slow->data);
slow = slow->ptr;
}
while (head != NULL )
{
int i=s.top();
s.pop();
if (head -> data != i)
{
return false ;
}
head=head->ptr;
}
return true ;
}
int main()
{
Node one = Node(1);
Node two = Node(2);
Node three = Node(3);
Node four = Node(2);
Node five = Node(1);
five.ptr = NULL;
one.ptr = &two;
two.ptr = &three;
three.ptr = &four;
four.ptr = &five;
Node* temp = &one;
int result = isPalin(&one);
if (result == 1)
cout << "isPalindrome is true" ;
else
cout << "isPalindrome is true" ;
return 0;
}
|
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 contains 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’.
C++
#include <bits/stdc++.h>
using namespace std;
struct Node
{
char data;
struct Node* next;
};
void reverse( struct Node**);
bool compareLists( struct Node*,
struct Node*);
bool isPalindrome( struct Node* head)
{
struct Node *slow_ptr = head,
*fast_ptr = head;
struct Node *second_half,
*prev_of_slow_ptr = head;
struct Node* midnode = NULL;
bool res = true ;
if (head != NULL &&
head->next != NULL)
{
while (fast_ptr != NULL &&
fast_ptr->next != NULL)
{
fast_ptr = fast_ptr->next->next;
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr->next;
}
if (fast_ptr != NULL)
{
midnode = slow_ptr;
slow_ptr = slow_ptr->next;
}
second_half = slow_ptr;
prev_of_slow_ptr->next = NULL;
reverse(&second_half);
res = compareLists(head, second_half);
reverse(&second_half);
if (midnode != NULL)
{
prev_of_slow_ptr->next = midnode;
midnode->next = second_half;
}
else
prev_of_slow_ptr->next = second_half;
}
return res;
}
void reverse( struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
struct Node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
bool compareLists( struct Node* head1,
struct Node* head2)
{
struct Node* temp1 = head1;
struct Node* temp2 = head2;
while (temp1 && temp2)
{
if (temp1->data == temp2->data)
{
temp1 = temp1->next;
temp2 = temp2->next;
}
else
return 0;
}
if (temp1 == NULL && temp2 == NULL)
return 1;
return 0;
}
void push( struct Node** head_ref,
char new_data)
{
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct Node* ptr)
{
while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" << "" ;
}
int main()
{
struct Node* head = NULL;
char str[] = "abacaba" ;
int i;
for (i = 0; str[i] != '' ; i++)
{
push(&head, str[i]);
printList(head);
isPalindrome(head) ? cout << "Is Palindrome" <<
"" : cout << "Not Palindrome" << "" ;
}
return 0;
}
|
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)
METHOD 3 (Using Recursion):
Use two pointers left and right. Move right and left using recursion and check for following in each recursive call.
1) Sub-list is a palindrome.
2) Value at current left and right are matching.
If both above conditions are true then return true.
The idea is to use function call stack as a container. Recursively traverse till the end of list. When we return from last NULL, we will be at the last node. The last node to be compared with first node of list.
In order to access first node of list, we need list head to be available in the last call of recursion. Hence, we pass head also to the recursive function. If they both match we need to compare (2, n-2) nodes. Again when recursion falls back to (n-2)nd node, we need reference to 2nd node from the head. We advance the head pointer in the previous call, to refer to the next node in the list.
However, the trick is identifying a double-pointer. Passing a single pointer is as good as pass-by-value, and we will pass the same pointer again and again. We need to pass the address of the head pointer for reflecting the changes in parent recursive calls.
Thanks to Sharad Chandra for suggesting this approach.
C++
#include <bits/stdc++.h>
using namespace std;
struct node
{
char data;
struct node* next;
};
bool isPalindromeUtil( struct node** left,
struct node* right)
{
if (right == NULL)
return true ;
bool isp = isPalindromeUtil(left,
right->next);
if (isp == false )
return false ;
bool isp1 = (right->data == (*left)->data);
*left = (*left)->next;
return isp1;
}
bool isPalindrome( struct node* head)
{
isPalindromeUtil(&head, head);
}
void push( struct node** head_ref,
char new_data)
{
struct node* new_node =
( struct node*) malloc ( sizeof ( struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct node* ptr)
{
while (ptr != NULL)
{
cout << ptr->data << "->" ;
ptr = ptr->next;
}
cout << "NULL" ;
}
int main()
{
struct node* head = NULL;
char str[] = "abacaba" ;
int i;
for (i = 0; str[i] != '' ; i++)
{
push(&head, str[i]);
printList(head);
isPalindrome(head) ? cout <<
"Is Palindrome" : cout << "Not Palindrome" ;
}
return 0;
}
|
Output:
a->NULL
Not 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(n) if Function Call Stack size is considered, otherwise O(1).
Please refer complete article on Function to check if a singly linked list is palindrome for more details!
Last Updated :
11 Apr, 2023
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