Write an algorithm to check whether the given string is palindrome or not using linked list

Function 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.

Nội dung chính Show

• Function to check if a singly linked list is palindrome
• Program to determine whether a singly linked list is the palindrome
• Explanation
• Using Stack:
• Traversing the list:
• The following code implements the above algorithm:
• Check if a Linked List is Palindrome (C++ Code)
• Check if a Singly Linked List is a Palindrome-Interview Problem
• Problem Statement
• Video liên quan

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 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 not bool isPalin(Node* head){ // Temp pointer Node* slow= head; // Declare a stack stack 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 Code int 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\n"; else cout<<"isPalindrome is true\n"; return 0; } // This code has been contributed by Striver

Java

/* Java program to check if linked list is palindrome recursively */ import java.util.*; class linkeList { public static void main(String args[]) { Node one = new Node(1); Node two = new Node(2); Node three = new Node(3); Node four = new Node(4); Node five = new Node(3); Node six = new Node(2); Node seven = new Node(1); one.ptr = two; two.ptr = three; three.ptr = four; four.ptr = five; five.ptr = six; six.ptr = seven; boolean condition = isPalindrome(one); System.out.println("isPalidrome :" + condition); } static boolean isPalindrome(Node head) { Node slow = head; boolean ispalin = true; Stack stack = new Stack(); while (slow != null) { stack.push(slow.data); slow = slow.ptr; } while (head != null) { int i = stack.pop(); if (head.data == i) { ispalin = true; } else { ispalin = false; break; } head = head.ptr; } return ispalin; } } class Node { int data; Node ptr; Node(int d) { ptr = null; data = d; } }

Python3

# 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

C#

// C# program to check if linked list // is palindrome recursively using System; using System.Collections.Generic; class linkeList{ // Driver code public static void Main(String []args) { Node one = new Node(1); Node two = new Node(2); Node three = new Node(3); Node four = new Node(4); Node five = new Node(3); Node six = new Node(2); Node seven = new Node(1); one.ptr = two; two.ptr = three; three.ptr = four; four.ptr = five; five.ptr = six; six.ptr = seven; bool condition = isPalindrome(one); Console.WriteLine("isPalidrome :" + condition); } static bool isPalindrome(Node head) { Node slow = head; bool ispalin = true; Stack stack = new Stack(); while (slow != null) { stack.Push(slow.data); slow = slow.ptr; } while (head != null) { int i = stack.Pop(); if (head.data == i) { ispalin = true; } else { ispalin = false; break; } head = head.ptr; } return ispalin; } } class Node { public int data; public Node ptr; public Node(int d) { ptr = null; data = d; } } // This code is contributed by amal kumar choubey

Javascript

Output

isPalindrome: true

Time complexity: O(n).

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’.

C++

// C++ program to check if a linked list is palindrome #include using namespace std; // Link list node struct 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 not bool 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_ptr 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(&second_half); // Reverse the second half again // 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 head 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; } // Function to check if two input // lists have same data 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; } // 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 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 list void printList(struct Node* ptr) { while (ptr != NULL) { cout << ptr->data << "->"; ptr = ptr->next; } cout << "NULL" << "\n"; } // Driver code int main() { // Start with the empty list struct Node* head = NULL; char str[] = "abacaba"; int i; for(i = 0; str[i] != '\0'; i++) { push(&head, str[i]); printList(head); isPalindrome(head) ? cout << "Is Palindrome" << "\n\n" : cout << "Not Palindrome" << "\n\n"; } return 0; } // This code is contributed by Shivani

C

/* Program to check if a linked list is palindrome */ #include #include #include /* Link list node */ struct 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 not */ 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; // To handle odd size list bool res = true; // initialize result if (head != NULL && head->next != NULL) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr 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; prev_of_slow_ptr->next = NULL; // NULL terminate first half reverse(&second_half); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(&second_half); // Reverse the second half again // 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 head */ 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; } /* Function to check if two input lists have same data*/ 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; } /* 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 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 pochar to the new node */ (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct Node* ptr) { while (ptr != NULL) { printf("%c->", ptr->data); ptr = ptr->next; } printf("NULL\n"); } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct Node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); printList(head); isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n"); } return 0; }

Java

/* Java program to check if linked list is palindrome */ class LinkedList { Node head; // head of list Node slow_ptr, fast_ptr, second_half; /* Linked list Node*/ class Node { char data; Node next; Node(char d) { data = d; next = null; } } /* Function to check if given linked list is palindrome or not */ boolean isPalindrome(Node head) { slow_ptr = head; fast_ptr = head; Node prev_of_slow_ptr = head; Node midnode = null; // To handle odd size list boolean res = true; // initialize result if (head != null && head.next != null) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr 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 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 != null) { midnode = slow_ptr; slow_ptr = slow_ptr.next; } // Now reverse the second half and compare it with first half second_half = slow_ptr; prev_of_slow_ptr.next = null; // NULL terminate first half reverse(); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(); // Reverse the second half again if (midnode != null) { // 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 */ void reverse() { Node prev = null; Node current = second_half; Node next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } second_half = prev; } /* Function to check if two input lists have same data*/ boolean compareLists(Node head1, Node head2) { Node temp1 = head1; Node temp2 = head2; while (temp1 != null && temp2 != null) { if (temp1.data == temp2.data) { temp1 = temp1.next; temp2 = temp2.next; } else return false; } /* Both are empty return 1*/ if (temp1 == null && temp2 == null) return true; /* Will reach here when one is NULL and other is not */ return false; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list off the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { System.out.print(ptr.data + "->"); ptr = ptr.next; } System.out.println("NULL"); } /* Driver program to test the above functions */ public static void main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char str[] = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; String string = new String(str); for (int i = 0; i < 7; i++) { llist.push(str[i]); llist.printList(llist.head); if (llist.isPalindrome(llist.head) != false) { System.out.println("Is Palindrome"); System.out.println(""); } else { System.out.println("Not Palindrome"); System.out.println(""); } } } }

Python3

# Python3 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 pallindrome 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_ptr 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 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 off the new one new_node.next = self.head # Move the head to point to 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\n") else: print("Not Palindrome\n") print() # This code is contributed by MuskanKalra1

C#

/* C# program to check if linked list is palindrome */ using System; class LinkedList { Node head; // head of list Node slow_ptr, fast_ptr, second_half; /* Linked list Node*/ public class Node { public char data; public Node next; public Node(char d) { data = d; next = null; } } /* Function to check if given linked list is palindrome or not */ Boolean isPalindrome(Node head) { slow_ptr = head; fast_ptr = head; Node prev_of_slow_ptr = head; Node midnode = null; // To handle odd size list Boolean res = true; // initialize result if (head != null && head.next != null) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr 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 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 != null) { midnode = slow_ptr; slow_ptr = slow_ptr.next; } // Now reverse the second half and compare it with first half second_half = slow_ptr; prev_of_slow_ptr.next = null; // NULL terminate first half reverse(); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(); // Reverse the second half again if (midnode != null) { // 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 */ void reverse() { Node prev = null; Node current = second_half; Node next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } second_half = prev; } /* Function to check if two input lists have same data*/ Boolean compareLists(Node head1, Node head2) { Node temp1 = head1; Node temp2 = head2; while (temp1 != null && temp2 != null) { if (temp1.data == temp2.data) { temp1 = temp1.next; temp2 = temp2.next; } else return false; } /* Both are empty return 1*/ if (temp1 == null && temp2 == null) return true; /* Will reach here when one is NULL and other is not */ return false; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list off the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { Console.Write(ptr.data + "->"); ptr = ptr.next; } Console.WriteLine("NULL"); } /* Driver program to test the above functions */ public static void Main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; for (int i = 0; i < 7; i++) { llist.push(str[i]); llist.printList(llist.head); if (llist.isPalindrome(llist.head) != false) { Console.WriteLine("Is Palindrome"); Console.WriteLine(""); } else { Console.WriteLine("Not Palindrome"); Console.WriteLine(""); } } } } // This code is contributed by Arnab Kundu

Javascript

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 using namespace std; /* Link list node */ struct node { char data; struct node* next; }; // Initial parameters to this function are &head and head bool 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 list void printList(struct node* ptr) { while (ptr != NULL) { cout << ptr->data << "->"; ptr = ptr->next; } cout << "NULL\n" ; } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); printList(head); isPalindrome(head) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n"; } return 0; } //this code is contributed by shivanisinghss2110

C

// Recursive program to check if a given linked list is palindrome #include #include #include /* Link list node */ struct node { char data; struct node* next; }; // Initial parameters to this function are &head and head bool 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 pochar to the new node */ (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct node* ptr) { while (ptr != NULL) { printf("%c->", ptr->data); ptr = ptr->next; } printf("NULL\n"); } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); printList(head); isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n"); } return 0; }

Java

// Java program for the above approach public class LinkedList{ // Head of the list Node head; Node left; public class Node { public char data; public Node next; // Linked list node public Node(char d) { data = d; next = null; } } // Initial parameters to this function are // &head and head boolean isPalindromeUtil(Node right) { left = head; // Stop recursion when right becomes null if (right == null) return true; // If sub-list is not palindrome then no need to // check for the current left and right, return // false boolean isp = isPalindromeUtil(right.next); if (isp == false) return false; // Check values at current left and right boolean isp1 = (right.data == left.data); left = left.next; // Move left to next node; return isp1; } // A wrapper over isPalindrome(Node head) boolean isPalindrome(Node head) { boolean result = isPalindromeUtil(head); return result; } // Push a node to linked list. Note that // this function changes the head public void push(char new_data) { // Allocate the node and put in the data Node new_node = new Node(new_data); // Link the old list off the the new one new_node.next = head; // Move the head to point to new node head = new_node; } // A utility function to print a // given linked list void printList(Node ptr) { while (ptr != null) { System.out.print(ptr.data + "->"); ptr = ptr.next; } System.out.println("Null"); } // Driver Code public static void main(String[] args) { LinkedList llist = new LinkedList(); char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; for(int i = 0; i < 7; i++) { llist.push(str[i]); llist.printList(llist.head); if (llist.isPalindrome(llist.head)) { System.out.println("Is Palindrome"); System.out.println(""); } else { System.out.println("Not Palindrome"); System.out.println(""); } } } } // This code is contributed by abhinavjain194

Python3

# Python program for the above approach # Head of the list head = None left = None class Node: def __init__(self, val): self.data = val self.next = None # Initial parameters to this function are # &head and head def isPalindromeUtil(right): global head, left left = head # Stop recursion when right becomes null if (right == None): return True # If sub-list is not palindrome then no need to # check for the current left and right, return # false isp = isPalindromeUtil(right.next) if (isp == False): return False # Check values at current left and right isp1 = (right.data == left.data) left = left.next # Move left to next node; return isp1 # A wrapper over isPalindrome(Node head) def isPalindrome(head): result = isPalindromeUtil(head) return result # Push a node to linked list. Note that # this function changes the head def push(new_data): global head # Allocate the node and put in the data new_node = Node(new_data) # Link the old list off the the new one new_node.next = head # Move the head to point to new node head = new_node # A utility function to print a # given linked list def printList(ptr): while (ptr != None): print(ptr.data, end="->") ptr = ptr.next print("Null ") # Driver Code str = ['a', 'b', 'a', 'c', 'a', 'b', 'a'] for i in range(0, 7): push(str[i]) printList(head) if (isPalindrome(head) and i != 0): print("Is Palindrome\n") else: print("Not Palindrome\n") # This code is contributed by saurabh_jaiswal.

C#

/* C# program to check if linked list is palindrome recursively */ using System; public class LinkedList { Node head; // head of list Node left; /* Linked list Node*/ public class Node { public char data; public Node next; public Node(char d) { data = d; next = null; } } // Initial parameters to this function are &head and head Boolean isPalindromeUtil(Node right) { left = head; /* 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 */ Boolean isp = isPalindromeUtil(right.next); if (isp == false) return false; /* Check values at current left and right */ Boolean isp1 = (right.data == (left).data); /* Move left to next node */ left = left.next; return isp1; } // A wrapper over isPalindromeUtil() Boolean isPalindrome(Node head) { Boolean result = isPalindromeUtil(head); return result; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list off the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { Console.Write(ptr.data + "->"); ptr = ptr.next; } Console.WriteLine("NULL"); } /* Driver code */ public static void Main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char []str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; //String string = new String(str); for (int i = 0; i < 7; i++) { llist.push(str[i]); llist.printList(llist.head); if (llist.isPalindrome(llist.head) != false) { Console.WriteLine("Is Palindrome"); Console.WriteLine(""); } else { Console.WriteLine("Not Palindrome"); Console.WriteLine(""); } } } } // This code is contributed by Rajput-Ji

Javascript

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).

Article Tags :

Linked List

Accolite

Adobe

Amazon

KLA Tencor

Kritikal Solutions

Microsoft

palindrome

Snapdeal

Yodlee Infotech

Practice Tags :

Accolite

Amazon

Microsoft

Snapdeal

Adobe

Yodlee Infotech

KLA Tencor

Kritikal Solutions

Linked List

palindrome

Program to determine whether a singly linked list is the palindrome

Explanation

In this program, we need to check whether given singly linked list is a palindrome or not. A palindromic list is the one which is equivalent to the reverse of itself.

The list given in the above figure is a palindrome since it is equivalent to its reverse list, i.e., 1, 2, 3, 2, 1. To check whether a list is a palindrome, we traverse the list and check if any element from the starting half doesn't match with any element from the ending half, then we set the variable flag to false and break the loop.

In the last, if the flag is false, then the list is palindrome otherwise not. The algorithm to check whether a list is a palindrome or not is given below.

Algorithm

1. Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
2. Create another class Palindrome which has three attributes: head, tail, and size.
3. addNode() will add a new node to the list:
   1. Create a new node.
   2. It first checks, whether the head is equal to null which means the list is empty.
   3. If the list is empty, both head and tail will point to a newly added node.
   4. If the list is not empty, the new node will be added to end of the list such that tail's next will point to a newly added node. This new node will become the new tail of the list.
4. reverseList() will reverse the order of the node present in the list:
   1. Node current will represent a node from which a list needs to be reversed.
   2. Node prevNode represent the previous node to current and nextNode represent the node next to current.
   3. The list will be reversed by swapping the prevNode with nextNode for each node.
5. isPalindrome() will check whether given list is palindrome or not:
   1. Declare a node current which will initially point to head node.
   2. The variable flag will store a boolean value true.
   3. Calculate the mid-point of the list by dividing the size of the list by 2.
   4. Traverse through the list till current points to the middle node.
   5. Reverse the list after the middle node until the last node using reverseList(). This list will be the second half of the list.
   6. Now, compare nodes of first half and second half of the list.
   7. If any of the nodes don't match then, set a flag to false and break the loop.
   8. If the flag is true after the loop which denotes that list is a palindrome.
   9. If the flag is false, then the list is not a palindrome.
6. display() will display the nodes present in the list:
   1. Define a node current which will initially point to the head of the list.
   2. Traverse through the list till current points to null.
   3. Display each node by making current to point to node next to it in each iteration.

Solution

Python

Output:

C

Output:

JAVA

Output:

C#

Output:

PHP

Output:

Using Stack:

Algorithm

• Traverse the linked list and push the value in Stack.
• Now, again traverse the linked list and while doing so pop the values from the stack.If stack become empty it means given Linked list is pallindrom otherwise not.

Given below is a Singly Linked List:

Traversing the list:

Step 1:

Currently,pointer is at the beginning of the list which points to Node value 2

Step 2:

Now,After the Node 2 is visited pointer moves forward and the visited node will be push into the stack.

Step 3:

Node 3 is visited,push Node 3 into Stack and moves the pointer forward.

Step 4:

Node 3 is visited,push Node 3 into Stack and moves the pointer forward.

Step 5:

Node 2 is visited,push Node 2 into Stack and moves the pointer forward until it's NULL.

Step 6:

Now,our approach will be to traverse the list once again from the beginning and this time will be ke checking is the Stack.top() is equal to the current node or not.Here,top value of stack is 2 and current node value of list is 2 they both are equal then pop the top value from the stack and move the pointer forward,repeat these steps until list found to be NULL.
Whenever you found that top value of stack is not equal to the current value of node then return false.

Step 7:

Step 8:

Step 9:

Step 10:

The following code implements the above algorithm:

Every engineering student aspires to get a decent placement, but the journey to that goal isn’t very easy for most of us, don’t worry in this tutorial we are going to take a little load off your shoulders and discuss one of the most frequently asked interview questions on the topic linked list. This particular problem statement has been asked in technical interviews of many multinational companies like Google, Facebook, and Uber. Let’s understand first what exactly is a palindrome.

Check if a Singly Linked List is a Palindrome-Interview Problem

Difficulty: Medium

Asked in: Amazon, Microsoft, Adobe, Flipkart

Understanding the Problem:

Given a singly Linked List, you have to find out whether the linked list is a palindrome or not. A palindrome is nothing but a string or number which is same whether you read it from right to left or left to right. It is like the string and its mirror image both are same.

For example,

Input: 1->2->2->1->NULL

Output: Yes

Input: 7->7->7->NULL

Output: Yes

Input: 1->2->3->NULL

Output: No

Possible follow-up questions to ask the interviewer: →

• Is the data value of the given Linked List is only numbers? (Ans: No, the data values can be anything.)

Linked List’s Node Class

Solutions

We are going to discuss three different ways to approach this problem. Also, we will be writing the pseudo-codes for all the three possible solutions.

• Solution Using Stack: Here we will be using the stack data structure. Stack has a property that the element inserted first will be taken out last. This property will help us to find whether the given Linked List is palindrome or not.
• Solution By Reversing the Second Half: By reversing the second half, and making a little observation we can solve this problem easily.
• Solution with Recursion: We will here make use of recursion stack in a much smarter way to find our solution.

Solution Using Stack

Solution idea

The idea is very simple and straight-forward, we all know that stack is based on FILO(First In Last Out) principle. We will keep pushing all the elements of Linked List into the stack and once the linked list is fully traversed we will start popping the elements out of the stack and comparing with the linked list’s elements.

For better understanding the approach, let’s see an explanation: →

Initially, our stack S is empty, we will start pushing elements of the linked list one after the other:

Now we will start popping an element out of the stack and comparing it with the linked list’s data starting from the head.

Similarly, if all the elements from both stack and linked list get matched as in this case, we can conclude that the linked list is a palindrome.

Solution steps

• We will create a stack S and push all the elements of the Linked List into the stack.
• Then we will pop the elements one by one from the stack and compare it with the data present in the Linked List sequentially.
• If there is any mismatch in comparison, we can conclude that the Linked List is not a palindrome else it is a palindrome.

Pseudo-code

S ListNode temp = head // traversing the linked list and pushing the elements into stack while(temp is not NULL) { S.push(temp.data) temp=temp.next } temp = head // again traversing the linked list to compare while(temp is not NULL) { topOfStack = S.top() S.pop() if(topOfStack is not equal temp) return False temp=temp.next } return True }

Complexity Analysis

Time Complexity: O(L), where L is the length of the given linked list.

Space Complexity: How much extra space used? (Think!)

Critical ideas to think!

• Take an example of palindrome, divide the palindrome into two equal parts. Reverse the second half of the palindrome. Have you got any observation?

Solution By Reversing the Second Half

Solution idea

We will divide our Linked List into two halves by finding the middle node. When we reverse the second half of the Linked List and if the Linked List is a palindrome then the second half becomes exactly same to the first half of the Linked List. This solution is based on this idea.

Solution steps

• We will first find the middle node of the given Linked List(Consider both the odd and even cases).
• Then, we will reverse the second half of the Linked List.
• We will compare the second half with the first half, if both the halves are exactly the same then the linked list is a palindrome.
• Reconstruct the actual given Linked List by again reversing the second half and attaching it to the first half.

Pseudo-code

boolean compareList(firstHalf , secondHalf) { while(firstHalf is not NULL and secondHalf is not NULL) { if(firstHalf.data is not equal secondHalf.data) return False firstHalf = firstHalf.next secondHalf = secondHalf.next } return True } ListNode reverseList(ListNode head) { if(head is NULL or head.next is NULL) return head ListNode reversedPart = reverseList(head.next) head.next.next = head head.next = NULL return reversedPart } boolean isListPalindrome(ListNode head) { ListNode midNode = NULL ListNode slow_pointer = head ListNode previous_of_mid = NULL ListNode fast_pointer = head.next if(head is not NULL and head.next is not NULL) { while(fast_pointer is not NULL and fast_pointer.next is not NULL) { fast_pointer = fast_pointer.next.next previous_of_mid = slow_pointer slow_pointer = slow_pointer.next } if(fast_pointer is not NULL) //If there are odd nodes { midNode = slow_pointer slow_pointer = slow_pointer.next } ListNode secondHalf = slow_pointer ListNode reversedHalf = reverse(secondHalf) boolean result = compareList(head, reversedHalf) } }

Complexity Analysis

Time Complexity: O(N), where N is the length of the Linked List

Space Complexity: O(N) (Why? Can you reduce this complexity?)

Critical ideas to think!

• You must have got the idea of how to reduce the space complexity(Think if not). Learn more about the algorithm used here to find the middle node.

Solution Using Recursion

Solution idea

This solution uses the recursion call stack in a very efficient way. We will take two pointers, one for the leftmost node and the other one for the rightmost node. We will check if the data of the left and the right nodes matches or not. If there is a match, we will go on recursively to check for the rest of the list. At last, if all the calls return true (that conveys match), we can conclude that the given Linked List is a palindrome.

Solution steps

• Recursively traverse the entire linked list to get the last node as a rightmost node.
• Compare the first node (left node) with the right node. If both are having same data value then recursively call for the sub-list as to check if the sub-list is a palindrome or not.
• If all the recursive calls are returning true, it means the Linked List given is a palindrome else it is not a palindrome.

Pseudo-code

boolean isListPalindrome(ListNode &leftmostNode, ListNode rightmostNode) { if(rightmostNode == NULL) return True //check if the sub-list is a palindrome boolean resultSublist = isListPalindrome(leftmostNode, rightmostNode.next) if(resultSublist is False) //No need to check then return False boolean result = leftmostNode.data == rightmostNode.data leftmostNode = leftmostNode.next return result }

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(N) (Stack space)

Critical ideas to think!

• Try to understand the recursion call stack for this solution and getting a clear understanding of how recursion works.
• Here we have passed the left node pointer variable as a reference. Why?

Comparison of different Solutions

Suggested Problems to Solve

• Sum of the values present in the nodes of the Linked List.
• Convert a Linked List into an array.
• Check if the given Linked List is sorted or not.
• Convert a Circular Linked List into a Singly Linked List.

Happy Coding! Enjoy Algorithms!!