Palindrome Singly Linked List
O
package com.thealgorithms.datastructures.lists;
import java.util.Stack;
/**
* A simple way of knowing if a singly linked list is palindrome is to push all
* the values into a Stack and then compare the list to popped vales from the
* Stack.
*
* See more:
* https://www.geeksforgeeks.org/function-to-check-if-a-singly-linked-list-is-palindrome/
*
* @see com.thealgorithms.strings.Palindrome
* @see com.thealgorithms.stacks.PalindromeWithStack
* @see com.thealgorithms.bitmanipulation.BinaryPalindromeCheck
* @see com.thealgorithms.maths.LowestBasePalindrome
* @see com.thealgorithms.maths.PalindromePrime
* @see com.thealgorithms.maths.PalindromeNumber
*/
@SuppressWarnings("rawtypes")
public final class PalindromeSinglyLinkedList {
private PalindromeSinglyLinkedList() {
}
public static boolean isPalindrome(final Iterable linkedList) {
var linkedListValues = new Stack<>();
for (final var x : linkedList) {
linkedListValues.push(x);
}
for (final var x : linkedList) {
if (x != linkedListValues.pop()) {
return false;
}
}
return true;
}
// Optimised approach with O(n) time complexity and O(1) space complexity
public static boolean isPalindromeOptimised(Node head) {
if (head == null || head.next == null) {
return true;
}
Node slow = head;
Node fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
Node midNode = slow;
Node prevNode = null;
Node currNode = midNode;
Node nextNode;
while (currNode != null) {
nextNode = currNode.next;
currNode.next = prevNode;
prevNode = currNode;
currNode = nextNode;
}
Node left = head;
Node right = prevNode;
while (left != null && right != null) {
if (left.val != right.val) {
return false;
}
right = right.next;
left = left.next;
}
return true;
}
static class Node {
int val;
Node next;
Node(int val) {
this.val = val;
this.next = null;
}
}
}