[234]回文链表

//请判断一个链表是否为回文链表。 
//
// 示例 1: 
//
// 输入: 1->2
//输出: false 
//
// 示例 2: 
//
// 输入: 1->2->2->1
//输出: true
// 
//
// 进阶： 
//你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？ 
// Related Topics 链表 双指针 
// 👍 879 👎 0

/*
* 234 回文链表
* 2021-03-04 12:48:36
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 Solution2 {
public:
    bool isPalindrome(ListNode* head) {
        if(!head || !head->next) return true;
        stack<int> stk;
        auto ptr = head;
        while(ptr){
            stk.push(ptr->val);
            ptr = ptr->next;
        }

        while(head){
            if(head->val == stk.top())
                head = head->next, stk.pop();
            else
                return false;
        }
        return true;
    }
};

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(!head || !head->next) return false;
        auto slow = head, fast = head;

        while(fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }
        auto last = slow->next , pre = head;

        while(last->next){
            auto temp = last->next;
            last->next = temp->next;
            temp->next = slow->next;
            slow->next = temp;
        }

        while(slow->next){
            slow = slow->next;
            if(pre->val != slow->val) return false;
            pre = pre->next;
        }
        return true;
    }
};
//leetcode submit region end(Prohibit modification and deletion)