C++ Program To Check If A Singly Linked List Is Palindrome

15 Jun, 2022
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++
// C++ program to implement// the above approach#include<bits/stdc++.h>using namespace std; class Node{    public:    int data;    Node(int d)    {        data = d;    }    Node *ptr;}; // Function to check if the linked list// is palindrome or notbool isPalin(Node* head){           // Temp pointer    Node* slow= head;     // Declare a stack    stack <int> s;     // Push all elements of the list    // to the stack    while(slow != NULL)    {        s.push(slow->data);         // Move ahead        slow = slow->ptr;    }     // Iterate in the list again and    // check by popping from the stack    while(head != NULL )    {                   // Get the top most element        int i=s.top();         // Pop the element        s.pop();         // Check if data is not        // same as popped element        if(head -> data != i)        {            return false;        }         // Move ahead        head=head->ptr;        }        return true;} // Driver Codeint main(){    // Addition of linked list    Node one =  Node(1);    Node two = Node(2);    Node three = Node(3);    Node four = Node(2);    Node five = Node(1);     // Initialize the next pointer    // of every current pointer    five.ptr = NULL;    one.ptr = &two;    two.ptr = &three;    three.ptr = &four;    four.ptr = &five;    Node* temp = &one;          // Call function to check    // palindrome or not    int result = isPalin(&one);       if(result == 1)            cout << "isPalindrome is true";    else        cout << "isPalindrome is true";     return 0;}// This code has been contributed by Striver
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++
// C++ program to check if a linked list// is palindrome#include <bits/stdc++.h>using namespace std; // Link list nodestruct Node{    char data;    struct Node* next;}; void reverse(struct Node**);bool compareLists(struct Node*,                  struct Node*); // Function to check if given linked list// is palindrome or notbool isPalindrome(struct Node* head){    struct Node *slow_ptr = head,                *fast_ptr = head;    struct Node *second_half,                *prev_of_slow_ptr = head;         // To handle odd size list    struct Node* midnode = NULL;         // initialize result    bool res = true;     if (head != NULL &&        head->next != NULL)    {                 // 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 != NULL &&               fast_ptr->next != NULL)        {            fast_ptr = fast_ptr->next->next;             // We need previous of the slow_ptr            // for linked lists with odd elements            prev_of_slow_ptr = slow_ptr;            slow_ptr = slow_ptr->next;        }         // fast_ptr would become NULL when there        // are even elements in 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 != NULL)        {            midnode = slow_ptr;            slow_ptr = slow_ptr->next;        }         // Now reverse the second half and        // compare it with first half        second_half = slow_ptr;                 // NULL terminate first half        prev_of_slow_ptr->next = NULL;                 // Reverse the second half        reverse(&second_half);                 // compare        res = compareLists(head, second_half);         // Construct the original list back        // Reverse the second half again        reverse(&second_half);         // If there was a mid node (odd size case)        // which was not part of either first half        // or 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;} // Function to reverse the linked list// Note that this function may change// the headvoid 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;} // Function to check if two input// lists have same databool 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;    }     // Both are empty return 1    if (temp1 == NULL && temp2 == NULL)        return 1;     // Will reach here when one is NULL    // and other is not    return 0;} // Push a node to linked list. Note// that this function changes the headvoid push(struct Node** head_ref,          char new_data){        // Allocate node    struct Node* new_node =          (struct Node*)malloc(sizeof(struct Node));     // Put in the data    new_node->data = new_data;     // Link the old list off the new node    new_node->next = (*head_ref);     // Move the head to point to the new node    (*head_ref) = new_node;} // A utility function to print a// given linked listvoid printList(struct Node* ptr){    while (ptr != NULL)    {        cout << ptr->data << "->";        ptr = ptr->next;    }    cout << "NULL" << "";} // Driver codeint main(){        // Start with the empty list    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;}// This code is contributed by Shivani
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++
// Recursive program to check if a given// linked list is palindrome#include <bits/stdc++.h>using namespace std; // Link list nodestruct node{    char data;    struct node* next;}; // Initial parameters to this function// are &head and headbool isPalindromeUtil(struct node** left,                      struct node* right){    // Stop recursion when right    // becomes NULL    if (right == NULL)        return true;     /* If sub-list is not palindrome then no       need to check for current left and right,       return false */    bool isp = isPalindromeUtil(left,                                right->next);    if (isp == false)        return false;     // Check values at current left and right    bool isp1 = (right->data == (*left)->data);     // Move left to next node    *left = (*left)->next;     return isp1;} // A wrapper over isPalindromeUtil()bool isPalindrome(struct node* head){    isPalindromeUtil(&head, head);} /* Push a node to linked list. Note that   this function changes the head */void push(struct node** head_ref,          char new_data){    // Allocate node    struct node* new_node =          (struct node*)malloc(sizeof(struct node));     // Put in the data    new_node->data = new_data;     // Link the old list off the new node    new_node->next = (*head_ref);     // Move the head to point to the new node    (*head_ref) = new_node;} // A utility function to print a// given linked listvoid printList(struct node* ptr){    while (ptr != NULL)    {        cout << ptr->data << "->";        ptr = ptr->next;    }    cout << "NULL" ;} // Driver codeint main(){    // Start with the empty list    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;}// This code is contributed by shivanisinghss2110
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! 
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