---
tags: leetcode
---

# [234. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)
---
# My Solution
## Solution 1: Stack
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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) {
        if (head == nullptr || head->next == nullptr) {
            return true;
        }

        stack<int> st;
        for (ListNode *it = head; it != nullptr; it = it->next) {
            st.push(it->val);
        }

        for (ListNode *it = head; it != nullptr; it = it->next) {
            if (st.top() != it->val) {
                return false;
            }
            st.pop();
        }
        return true;
    }
};

### Time Complexity
$O(n)$
* $n$ is the number of nodes in the linked list.
### Space Complexity
$O(n)$
## Solution 2: Break the linked list
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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) {
        if (head == nullptr || head->next == nullptr) {
            return true;
        }

        ListNode *preSlow = nullptr, *slow = head, *fast = head;
        while (fast != nullptr && fast->next != nullptr) {
            preSlow = slow;
            slow = slow->next;
            fast = fast->next->next;
        }

        ListNode *tmpHead = slow;
        preSlow->next = nullptr;

        ListNode *head2 = nullptr;
        while (tmpHead != nullptr) {
            ListNode *tmp = tmpHead->next;
            tmpHead->next = head2;
            head2 = tmpHead;
            tmpHead = tmp;
        }

        while (head != nullptr) {
            if (head->val != head2->val) {
                return false;
            }
            head = head->next;
            head2 = head2->next;
        }

        return true;
    }
};

### Time Complexity
$O(n)$
* $n$ is the number of nodes in the linked list.
### Space Complexity
$O(1)$
# Miscellaneous
<!--
# Test Cases

[1,2,2,1]

[1,2]

[3,3]

[1]

[1,2,4,2,1]

[1,2,4,1,2]

[1,2,1,2]

-->