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 :
Java
// Java program to check if linked list // is palindrome recursively import java.util.*; class linkedList { 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("isPalindrome :" + condition); } static boolean isPalindrome(Node head) { Node slow = head; boolean ispalin = true; Stack<Integer> stack = new Stack<Integer>(); 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; }} |
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 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’.
Java
// Java program to check if linked list// is palindrome class LinkedList { // Head of list Node head; 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; // To handle odd size list Node midnode = null; // Initialize result boolean 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 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; // NULL terminate first half prev_of_slow_ptr.next = null; // Reverse the second half reverse(); // compare res = compareLists(head, second_half); // Construct the original list back // Reverse the second half again reverse(); 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 of 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) { // 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(""); } } }} |
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 the list. When we return from the last NULL, we will be at the last node. The last node is to be compared with the first node of the list.
In order to access the first node of the list, we need the 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 a 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.
Java
// Java program to implement// the above approachpublic 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 of 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 |
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!
