C++ Program For Checking Linked List With A Loop Is Palindrome Or Not

Given a linked list with a loop, the task is to find whether it is palindrome or not. You are not allowed to remove the loop.  

Examples:

Input: 1 -> 2 -> 3 -> 2
             /|      |/
               ------- 1  
Output: Palindrome
Linked list is 1 2 3 2 1 which is a 
palindrome.

Input: 1 -> 2 -> 3 -> 4
             /|      |/
               ------- 1  
Output: Not Palindrome
Linked list is 1 2 3 4 1 which is a 
not palindrome. 

Algorithm:

1. Detect the loop using the Floyd Cycle Detection Algorithm.
2. Then find the starting node of the loop as discussed in this.
3. Check the linked list is palindrome or not as discussed in this.

Below is the implementation. 

C++

// C++ program to check if a linked list
// with loop is palindrome or not.
#include<bits/stdc++.h>
using namespace std;

// Link list node
struct Node
{
    int data;
    struct Node * next;
};

/* Function to find loop starting node.
   loop_node --> Pointer to one of the
   loop nodes head --> Pointer to the
   start node of the linked list */
Node* getLoopstart(Node *loop_node,
                   Node *head)
{
    Node *ptr1 = loop_node;
    Node *ptr2 = loop_node;

    // Count the number of nodes in
    // loop
    unsigned int k = 1, i;
    while (ptr1->next != ptr2)
    {
        ptr1 = ptr1->next;
        k++;
    }

    // Fix one pointer to head
    ptr1 = head;

    // And the other pointer to k nodes
    // after head
    ptr2 = head;
    for (i = 0; i < k; i++)
        ptr2 = ptr2->next;

    /* Move both pointers at the same pace,
       they will meet at loop starting node */
    while (ptr2 != ptr1)
    {
        ptr1 = ptr1->next;
        ptr2 = ptr2->next;
    }
    return ptr1;
}

/* This function detects and find loop
   starting node  in the list*/
Node* detectAndgetLoopstarting(Node *head)
{
    Node *slow_p = head, *fast_p = head,
         *loop_start;

    // Start traversing list and
    // detect loop
    while (slow_p && fast_p &&
           fast_p->next)
    {
        slow_p = slow_p->next;
        fast_p = fast_p->next->next;

        /* If slow_p and fast_p meet then
           find the loop starting node*/
        if (slow_p == fast_p)
        {
            loop_start = getLoopstart(slow_p,
                                      head);
            break;
        }
    }

    // Return starting node of loop
    return loop_start;
}

// Utility function to check if a linked list
// with loop is palindrome with given starting
// point.
bool isPalindromeUtil(Node *head,
                      Node* loop_start)
{
    Node *ptr = head;
    stack<int> s;

    // Traverse linked list until last node
    // is equal to loop_start and store the
    // elements till start in a stack
    int count = 0;
    while (ptr != loop_start || count != 1)
    {
        s.push(ptr->data);
        if (ptr == loop_start)
            count = 1;
        ptr = ptr->next;
    }
    ptr = head;
    count = 0;

    // Traverse linked list until last node is
    // equal to loop_start second time
    while (ptr != loop_start || count != 1)
    {
        // Compare data of node with the top
        // of stack
        // If equal then continue
        if (ptr->data == s.top())
            s.pop();

        // Else return false
        else
            return false;

        if (ptr == loop_start)
            count = 1;
        ptr = ptr->next;
    }

    // Return true if linked list
    // is palindrome
    return true;
}

// Function to find if linked list
// is palindrome or not
bool isPalindrome(Node* head)
{
    // Find the loop starting node
    Node* loop_start =
          detectAndgetLoopstarting(head);

    // Check if linked list is palindrome
    return isPalindromeUtil(head,
                            loop_start);
}

Node *newNode(int key)
{
    Node *temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}

// Driver code
int main()
{
    Node *head = newNode(50);
    head->next = newNode(20);
    head->next->next = newNode(15);
    head->next->next->next =
    newNode(20);
    head->next->next->next->next =
    newNode(50);

    // Create a loop for testing
    head->next->next->next->next->next =
    head->next->next;

    isPalindrome(head)? cout <<
    "Palindrome" : cout << "
    Not Palindrome";

    return 0;
}

Output:

Palindrome

Time Complexity: O(n) where n is no of nodes in the linked list

Auxiliary Space: O(n) where n is size of stack

 Approach 2: Constant Space.

 Here’s the approach to check if a linked list is palindrome or not without using extra space:

• Find the middle node of the linked list using the slow and fast pointer approach.
• Reverse the second half of the linked list.
  
• Traverse both halves of the linked list (first half from start to middle, and second half from middle to end) and compare their data.
• If all the data is same, then the linked list is palindrome. Otherwise, it’s not palindrome.
  Here’s the C++ code implementing the above approach:

C++

#include<bits/stdc++.h>
using namespace std;

// Link list node
struct Node
{
    int data;
    struct Node * next;
};

/* Function to reverse a linked list
from the given node */
Node* reverseList(Node *head)
{
    Node *prev = NULL;
    Node *curr = head;
    Node *next;

    while(curr != NULL)
    {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }

    return prev;
}

// Function to check if a linked list is palindrome or not
bool isPalindrome(Node* head)
{
    // Find the middle node of the linked list
    Node *slow = head, *fast = head;
    while(fast != NULL && fast->next != NULL)
    {
        slow = slow->next;
        fast = fast->next->next;
    }

    // Reverse the second half of the linked list
    Node *rev_second_half = reverseList(slow);

    // Traverse both halves of the linked list and compare their data
    Node *p1 = head, *p2 = rev_second_half;
    while(p2 != NULL)
    {
        if(p1->data != p2->data)
            return false;

        p1 = p1->next;
        p2 = p2->next;
    }

    return true;
}

// Utility function to create a new node
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}

// Driver code
int main()
{
    Node *head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(2);
    head->next->next->next->next = newNode(1);

    isPalindrome(head)? cout<<"Palindrome": cout<<"Not Palindrome";

    return 0;
}

Output

Palindrome

Time Complexity: O(n) where n is no of nodes in the linked list

Auxiliary Space: O(1) Uses no extra space.

Please refer complete article on Check linked list with a loop is palindrome or not for more details!