234. Palindrome Linked List

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234. Palindrome Linked List

Description

Difficulty: Easy

Related Topics: Linked List, Two Pointers, Stack, Recursion

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

Example 1:

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Input: head = [1,2,2,1]
Output: true

Example 2:

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Input: head = [1,2]
Output: false

Constraints:

  • The number of nodes in the list is in the range [1, 105].
  • 0 <= Node.val <= 9

Follow up: Could you do it in O(n) time and O(1) space?

Hints/Notes

  • Find the mid point, reverse the right part then compare

Solution

Language: C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:

    bool isPalindrome(ListNode* head) {
        ListNode* left = head, *slow = head, *fast = head;
        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        if (fast) {
            slow = slow->next;
        }
        ListNode* right = reverse(slow);
        // running the while loop against right since the end of right
        // is nullptr while left would still have items to iterate on
        while (right) {
            if (left->val != right->val) {
                return false;
            } else {
                right = right->next;
                left = left->next;
            }
        }
        return true;
    }

    ListNode* reverse(ListNode* head) {
        ListNode *cur = head, *prev = nullptr;
        while (cur) {
            ListNode* tmp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
    }

    /* Run time O(N), Space O(N) solution
    ListNode* left;

    bool isPalindrome(ListNode* head) {
        left = head;
        return traverse(head);
    }

    bool traverse(ListNode* right) {
        if (right == nullptr) return true;
        bool res = traverse(right->next);
        res = res && left->val == right->val;
        left = left->next;
        return res;
    }
    */
};