Determining if a Linked List is Palindromic

This problem asks whether or not a singly linked list is palindromic, meaning that the data contained within the list is the same whether read forwards or backwards. An example input is provided, along with the expected output.


Given a singly linked list, determine whether or not the list is palindromic.
Example input:
1 -> 2 -> 3 -> 2 -> 1
Example output:


This solution is in C++. Our A.I. can create solutions in multiple languages.
by kayakane
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <numeric>
#include <utility>
#include <limits>
#include <time.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>

using namespace std;

struct Node {
    int data;
    Node* next;

bool isPalindrome(Node* head) {
    Node* slow = head;
    Node* fast = head;
    stack<int> s;
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next; fast = fast->next->next;
    if (fast != NULL) {
        slow = slow->next;
    while (slow != NULL) {
        int top =;
        if (top != slow->data) {
            return false;
        slow = slow->next;
    return true;

int main() {
    Node* head = new Node();
    head->data = 1;
    head->next = new Node();
    head->next->data = 2;
    head->next->next = new Node();
    head->next->next->data = 3;
    head->next->next->next = new Node();
    head->next->next->next->data = 2;
    head->next->next->next->next = new Node();
    head->next->next->next->next->data = 1;
    cout << isPalindrome(head) << endl;
    return 0;

A.I. Evaluation of the Solution

The candidate's solution correctly determines whether or not a singly linked list is palindromic. The candidate uses a fast and slow pointer to reach the middle of the list, and then uses a stack to store the first half of the list. The candidate then compares the second half of the list to the values in the stack. This is a valid approach to solving the problem.

Evaluated at: 2022-11-23 10:16:13