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Solution: Palindrome Linked List

seanpgallivan
Fledgling software developer; the struggle is a Rational Approximation.
・4 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #234 (Easy): Palindrome Linked List


Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

Given the head of a singly linked list, return true if it is a palindrome.


Examples:

Example 1:
Input: head = [1,2,2,1]
Output: true
Visual: Example 1 Visual
Example 2:
Input: head = [1,2]
Output: false
Visual: Example 1 Visual

Constraints:

  • The number of nodes in the list is in the range [1, 10^5].
  • 0 <= Node.val <= 9

Idea:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

The naive approach here would be to run through the linked list and create an array of its values, then compare the array to its reverse to find out if it's a palindrome. Though this is easy enough to accomplish, we're challenged to find an approach with a space complexity of only O(1) while maintaining a time complexity of O(N).

The only way to check for a palindrome in O(1) space would require us to be able to access both nodes for comparison at the same time, rather than storing values for later comparison. This would seem to be a challenge, as the linked list only promotes travel in one direction.

But what if it didn't?

The answer is to reverse the back half of the linked list to have the next attribute point to the previous node instead of the next node. (Note: we could instead add a **prev* attribute as we iterate through the linked list, rather than overwriting next on the back half, but that would technically use O(N) extra space, just as if we'd created an external array of node values.*)

The first challenge then becomes finding the middle of the linked list in order to start our reversing process there. For that, we can look to Floyd's Cycle Detection Algorithm.

With Floyd's, we'll travel through the linked list with two pointers, one of which is moving twice as fast as the other. When the fast pointer reaches the end of the list, the slow pointer must then be in the middle.
Diagram 1
With slow now at the middle, we can reverse the back half of the list with the help of another variable to contain a reference to the previous node (prev) and a three-way swap. Before we do this, however, we'll want to set prev.next = null, so that we break the reverse cycle and avoid an endless loop.
Diagram 2
Once the back half is properly reversed and slow is once again at the end of the list, we can now start fast back over again at the head and compare the two halves simultaneously, with no extra space required.
Diagram 3
If the two pointers ever disagree in value, we can return false, otherwise we can return true if both pointers reach the middle successfully.

(Note: This process works regardless of whether the length of the linked list is odd or even, as the comparison will stop when slow reaches the "dead-end" node.)

Diagram 4


Implementation:

The code for all four languages is almost identical.


Javascript Code:


(Jump to: Problem Description || Solution Idea)

var isPalindrome = function(head) {
    let slow = head, fast = head, prev, temp
    while (fast && fast.next)
        slow = slow.next, fast = fast.next.next
    prev = slow, slow = slow.next, prev.next = null
    while (slow)
        temp = slow.next, slow.next = prev, prev = slow, slow = temp
    fast = head, slow = prev
    while (slow)
        if (fast.val !== slow.val) return false
        else fast = fast.next, slow = slow.next
    return true
};
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Python Code:


(Jump to: Problem Description || Solution Idea)

class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        slow, fast, prev = head, head, None
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        prev, slow, prev.next = slow, slow.next, None
        while slow:
            temp = slow.next
            slow.next, prev, slow = prev, slow, temp
        fast, slow = head, prev
        while slow:
            if fast.val != slow.val: return False
            fast, slow = fast.next, slow.next
        return True
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Java Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode slow = head, fast = head, prev, temp;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        prev = slow;
        slow = slow.next;
        prev.next = null;
        while (slow != null) {
            temp = slow.next;
            slow.next = prev;
            prev = slow;
            slow = temp;
        }
        fast = head;
        slow = prev;
        while (slow != null) {
            if (fast.val != slow.val) return false;
            fast = fast.next;
            slow = slow.next;
        }
        return true;
    }
}
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C++ Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode *slow = head, *fast = head, *prev, *temp;
        while (fast && fast->next)
            slow = slow->next, fast = fast->next->next;
        prev = slow, slow = slow->next, prev->next = NULL;
        while (slow)
            temp = slow->next, slow->next = prev, prev = slow, slow = temp;
        fast = head, slow = prev;
        while (slow)
            if (fast->val != slow->val) return false;
            else fast = fast->next, slow = slow->next;
        return true;
    }
};
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Discussion (2)

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chibisov profile image
Gennady Chibisov

Thank you for the article. I've built a bot which explains how to validate Palindrome Linked List bot.chib.me/courses/palindrome-lin...
Bot builds your intuition step by step without giving you a straight answer or solution.

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Vishal Gupta

Great Work Gennady. The Bot really helped me a lot in understanding the reverse linked list & palindrome linked list problem so clearly. Thank you so much again. 🔥🙌