Leetcode Palindrome Linked List problem solution

In this Leetcode Palindrome Linked List problem solution we have given the head of a singly linked list, return true if it is a palindrome.

Problem solution in Python.

class Solution(object):
    def isPalindrome(self, head):
        if not head:
            return True
        
        curr = head
        nums = []
        while curr:
            nums.append(curr.val)
            curr = curr.next
        
        left = 0
        right = len(nums) - 1
        
        while left <= right:
            if nums[left] != nums[right]:
                return False
            else:
                left += 1
                right -= 1
                
        return True

Problem solution in Java.

class Solution {
public boolean isPalindrome(ListNode head) {

    if (head == null || head.next == null) return true;
    
    Stack<Integer> stack = new Stack <Integer>();
    
    ListNode curr = head; 
    while (curr != null) {
        stack.push(curr.val);
        curr = curr.next;     
    }
   
    while (head != null) {
        if (stack.pop() != head.val) {
            return false;
        }
        else {
            head = head.next;
        }
    }
 
    return true;
}
}

Problem solution in C++.

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        vector<int>tab;
        ListNode* p;
        p=head;
        while(p!=nullptr){
            tab.push_back(p->val);
            p=p->next;
        }
        int n=tab.size();
        for(int i=0;i<n/2;i++){
            if(tab[i]!=tab[n-i-1])return false;
        }
        return true;
    }
};

Problem solution in C.

bool isPalindrome(struct ListNode* head){
    if(head==NULL){
        return true;
    }
    struct ListNode* p1=head;
    struct ListNode* p2=head->next;
    while(p2 && p2->next){
        p1 = p1->next;
        p2 = p2->next->next;
    }
    
    struct ListNode *prev, *curr, *n, *h2;
    prev = NULL;
    curr = p1->next;
    h2 = curr;
    while(curr){
        n = curr->next;
        curr->next = prev;
        prev = curr;
        curr = n;
    }
    p2 = head;
    p1 = prev;
    while(p1){
        if(p1->val!=p2->val){
            return false;
        }
        p1 = p1->next;
        p2 = p2->next;
    }
    
    return true;

}