Palindrome Linked List Solutions in C++

// Source : https://leetcode.com/problems/palindrome-linked-list/
// Author : Hao Chen
// Date   : 2015-07-17



/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* findMiddle(ListNode* head) {
        ListNode *p1=head, *p2=head;
        while(p2 && p2->next){
            p1 = p1->next;
            p2 = p2->next->next;
        }
        return p1;
    }
    
    ListNode* reverseLink(ListNode* head) {
        ListNode* p=NULL;
        
        while (head) {
            ListNode* q = head->next;
            head->next = p;
            p = head;
            head = q;
        }
        return p;
    }
    
    bool isPalindrome(ListNode* head) {
        ListNode* pMid = findMiddle(head);
        ListNode* pRev = reverseLink(pMid); 
        for(;head!=pMid; head=head->next, pRev=pRev->next) {
            if (head->val != pRev->val) {
                return false;
            }
        }
        return true;
    }
};

/// Source : https://leetcode.com/problems/palindrome-linked-list/description/
/// Author : liuyubobobo
/// Time   : 2018-05-10

#include <iostream>

using namespace std;

/// Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

/// Two Pointers to Reverse and Traverse
/// Time Complexity: O(n)
/// Space Complexity: O(1)
class Solution {
public:
    bool isPalindrome(ListNode* head) {

        if(head == NULL || head->next == NULL)
            return true;

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

        slow->next = reverse(slow->next);

        slow = slow->next;
        ListNode* cur = head;
        while(slow != NULL){
            if(cur->val != slow->val)
                return false;
            else{
                slow = slow->next;
                cur = cur->next;
            }
        }
        return true;
    }

private:
    ListNode* reverse(ListNode* head){

        if(head == NULL || head->next == NULL)
            return head;

        ListNode* pre = head;
        ListNode* cur = head->next;
        ListNode* next = cur->next;
        head->next = NULL;

        while(true){
            cur->next = pre;
            pre = cur;
            cur = next;
            if(cur == NULL)
                break;
            next = cur->next;
        }

        return pre;
    }
};

int main() {

    return 0;
}