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.
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++
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#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] coutnext; // 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 data next; } cout 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
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/* 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
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# 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 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_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#
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/* 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
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/* javascript program to check if linked list is palindrome */ var head; // head of list var slow_ptr, fast_ptr, second_half; /* Linked list Node */ class Node { constructor[val] { this.data = val; this.next = null; } } /* * Function to check if given linked list is palindrome or not */ function isPalindrome[head] { slow_ptr = head; fast_ptr = head; var prev_of_slow_ptr = head; var midnode = null; // To handle odd size list var 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 */ function reverse[] { var prev = null; var current = second_half; var 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 */ function compareLists[head1, head2] { var temp1 = head1; var 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 */ function push[ new_data] { /* * Allocate the Node & Put in the data */ var 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 function printList[ptr] { while [ptr != null] { document.write[ptr.data + "->"]; ptr = ptr.next; }
document.write["NULL } /* Driver program to test the above functions */
/* Start with the empty list */ var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]; var string = str.toString[]; for [i = 0; i < 7; i++] { push[str[i]]; printList[head]; if [isPalindrome[head] != false] { document.write["Is Palindrome"];
document.write[" } else { document.write["Not Palindrome"];
document.write[" } } // This code contributed by gauravrajput1 |
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 PalindromeTime 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++
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// 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 data next; } cout 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
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// 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
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# 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#
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/* 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
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// javascript program for the above approach // Head of the list var head; var left; class Node { constructor[val] { this.data = val; this.next = null; } } // Initial parameters to this function are // &head and head function isPalindromeUtil[ 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 var isp = isPalindromeUtil[right.next]; if [isp == false] return false; // Check values at current left and right var isp1 = [right.data == left.data]; left = left.next; // Move left to next node; return isp1; } // A wrapper over isPalindrome[Node head] function isPalindrome[ head] { var result = isPalindromeUtil[head]; return result; } // Push a node to linked list. Note that // this function changes the head function push[ new_data] { // Allocate the node and put in the data var 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 function printList[ ptr] { while [ptr != null] { document.write[ptr.data + "->"]; ptr = ptr.next; } document.write["Null "];
document.write[" } // Driver Code
var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]; for [var i = 0; i < 7; i++] { push[str[i]]; printList[head]; if [isPalindrome[head]] { document.write["Is Palindrome"];
document.write["
document.write[" } else { document.write["Not Palindrome"];
document.write["
document.write[" } }
// This code contributed by aashish2995 |
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 PalindromeTime Complexity: O[n]
Auxiliary Space: O[n] if Function Call Stack size is considered, otherwise O[1].
Article Tags :
Linked List
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palindrome
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Practice Tags :
Accolite
Amazon
Microsoft
Snapdeal
Adobe
Yodlee Infotech
KLA Tencor
Kritikal Solutions
Linked List
palindrome
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Palindrome Linked List in Python
PythonServer Side ProgrammingProgramming
Suppose we have a linked list. We have to check whether the list elements are forming a palindrome or not. So if the list element is like [1,2,3,2,1], then this is a palindrome.
To solve this, we will follow these steps −
fast := head, slow := head, rev := None and flag := 1
if the head is empty, then return true
while fast and next of fast is available
if next of the next of fast is available, then set flag := 0 and break the loop
fast := next of the next of fast
temp := slow, slow := next of slow
next of temp := rev, and rev := temp
fast := next of slow, and next of slow := rev
if flag is set, then slow := next of slow
while fast and slow are not None,
if the value of fast is not the same as the value of slow, then return false
fast := next of fast, and slow := next of slow
return true
C
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 | #include #include // A Linked List Node struct Node { int data; struct Node* next; }; // Helper function to create a new node with the given data and // pushes it onto the list's front void push[struct Node** head, int data] { // create a new linked list node from the heap struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; newNode->next = *head; *head = newNode; } // Recursive function to check if the linked list is a palindrome or not int checkPalindrome[struct Node** left, struct Node* right] { // base case if [right == NULL] { return 1; } int result = checkPalindrome[left, right->next] && [[*left]->data == right->data]; [*left] = [*left]->next; return result; } // Function to check if the linked list is a palindrome or not int checkPalin[struct Node* head] { return checkPalindrome[&head, head]; } int main[void] { // input keys int keys[] = { 1, 3, 5, 3, 1}; int n = sizeof[keys] / sizeof[keys[0]]; struct Node* head = NULL; for [int i = n - 1; i >= 0; i--] { push[&head, keys[i]]; } if [checkPalin[head]] { printf["The linked list is a palindrome"]; } else { printf["The linked list is not a palindrome"]; } return 0; } |
DownloadRun Code
Output:
The linked list is a palindrome
Palindrome LinkedList — Day 2[Python]
Today is my second day at the 365 days of coding challenge. I will be solving a question based on the LinkedList. Before I dive into the question let me talk about a few features of LinkedList.
- A LinkedList contains a sequence of nodes that are connected to each other using a pointer to the following nodes.
2. A basic node will contain 2 parts,
a. Data
b. Address for next pointer
3. A LinkedList has a head pointer that points towards the start of the LinkedList.
4. A LinkedList does not necessarily have a contiguous memory allocation.
The below code shows the structure of a node on a LinkedList.
class Node:def __init__[self, data]:
self.data = data
self.next = None
Okay, now that we have seen about LinkedList lets move on to today's question.
234. Palindrome Linked List
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2Output: false
Example 2:
Input: 1->2->2->1Output: true
My first thought after seeing this question was to use a stack, insert data from each node into the stack, and then compare elements by popping elements from the stack and each element in the linked list.
The code would look like the following
class Solution:def insert_stack[self, head]:
stack_rev = []
while[head]:
stack_rev.append[head.val]
head = head.next
return stack_rev
def isPalindrome[self, head: ListNode] -> bool:
if head == None:
return True
stack_rev = self.insert_stack[head]
while[head]:
if[head.val == stack_rev.pop[]]:
head = head.next
else:
return False
return True
Complexity analysis
Time Complexity
We would be required to traverse through each of the element in LinkedList and that would take O[N] time
Space Complexity
We are using a stack to store the elements which will take up O[N] space.
Can we try reducing space usage?
Maybe if we divide our LinkedList in the middle, reverse the first half and then compare the nodes in both the half.
class Solution:def isPalindrome[self, head]:
rev = None
# Slow pointer
slow = head
# Fast pointer
fast = head# Until the first half keep reversing the nodes
while fast and fast.next:
#Keep moving in double speed
fast = fast.next.next
#Keep reversing the node
slow_next = slow.next
slow.next = rev
rev = slow
slow = slow_next
#In case lenght of linkedlist is odd
if fast:
slow = slow.next
# Compare the elements
while rev and rev.val == slow.val:
slow = slow.next
rev = rev.next
return not rev
Complexity analysis
Time Complexity
We would be required to traverse through each of the element in LinkedList and that would take O[N] time
Space Complexity
We are not using any extra space in order to perform the above transformation and hence the space used is O[1].
I would like to improve my writing skills so any suggestions or critiques are highly welcomed.
Checking if a linked list is a palindrome
A number is a palindrome if it reads the same forwards as it does backward. For example, the number 123332112333211233321 is a palindrome.
Linked lists can also be palindromes if they have the same order of elements when traversed forwards and backward.
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
- Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
- Create another class Palindrome which has three attributes: head, tail, and size.
- addNode[] will add a new node to the list:
- Create a new node.
- It first checks, whether the head is equal to null which means the list is empty.
- If the list is empty, both head and tail will point to a newly added node.
- 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.
- reverseList[] will reverse the order of the node present in the list:
- Node current will represent a node from which a list needs to be reversed.
- Node prevNode represent the previous node to current and nextNode represent the node next to current.
- The list will be reversed by swapping the prevNode with nextNode for each node.
- isPalindrome[] will check whether given list is palindrome or not:
- Declare a node current which will initially point to head node.
- The variable flag will store a boolean value true.
- Calculate the mid-point of the list by dividing the size of the list by 2.
- Traverse through the list till current points to the middle node.
- Reverse the list after the middle node until the last node using reverseList[]. This list will be the second half of the list.
- Now, compare nodes of first half and second half of the list.
- If any of the nodes don't match then, set a flag to false and break the loop.
- If the flag is true after the loop which denotes that list is a palindrome.
- If the flag is false, then the list is not a palindrome.
- display[] will display the nodes present in the list:
- Define a node current which will initially point to the head of the list.
- Traverse through the list till current points to null.
- Display each node by making current to point to node next to it in each iteration.
Solution
Python
Output:
Nodes of the singly linked list: 1 2 3 2 1 Given singly linked list is a palindrome
C
Output:
Nodes of the singly linked list: 1 2 3 2 1 Given singly linked list is a palindrome
JAVA
Output:
Nodes of singly linked list: 1 2 3 2 1 Given singly linked list is a palindrome
C#
Output:
Nodes of singly linked list: 1 2 3 2 1 Given singly linked list is a palindrome
PHP
Output:
Nodes of singly linked list: 1 2 3 2 1 Given singly linked list is a palindrome