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This approach involves reversing the second half of the linked list and then comparing it with the first half. If they are identical, the linked list is a palindrome. The steps include:
Time Complexity: O(n) since we are traversing the list multiple times but each in linear time.
Space Complexity: O(1) as we are reversing the linked list in place.
1class ListNode {
2 int val;
3 ListNode next;
4 ListNode(int x) { val = x; }
5}
6
7public class Solution {
8 private ListNode reverseList(ListNode head) {
9 ListNode prev = null;
10 while (head != null) {
11 ListNode nextNode = head.next;
12 head.next = prev;
13 prev = head;
14 head = nextNode;
15 }
16 return prev;
17 }
18
19 public boolean isPalindrome(ListNode head) {
20 if (head == null || head.next == null) {
21 return true;
22 }
23 ListNode slow = head, fast = head;
24 while (fast != null && fast.next != null) {
25 slow = slow.next;
26 fast = fast.next.next;
27 }
28 ListNode secondHalf = reverseList(slow);
29 ListNode firstHalf = head;
30 while (secondHalf != null) {
31 if (firstHalf.val != secondHalf.val) {
32 return false;
33 }
34 firstHalf = firstHalf.next;
35 secondHalf = secondHalf.next;
36 }
37 return true;
38 }
39}Here, the Java solution involves reversing the second half of the linked list. The reversal and comparison steps efficiently determine if the list is a palindrome.
In this approach, convert the linked list into an array and utilize a two-pointer technique to determine if it forms a palindrome. The steps are as follows:
Time Complexity: O(n) due to single traversation and subsequent O(n/2) check.
Space Complexity: O(n) because we store all node values in an array.
This JavaScript implementation relies on an array to store linked list values. Crosschecking is done using two pointers, validating if input represents a palindromic state.