请判断一个链表是否为回文链表。
示例 1:

  1. 输入: 1->2
  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？

快慢指针

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode* slow = head, * fast = head;
        while(fast != NULL && fast->next != NULL){
            slow = slow->next;
            fast = fast->next->next;
        }
        if(fast != NULL){
            slow = slow->next;
        }
        slow = reverse(slow);

        fast = head;
        while(slow != NULL){
            if(slow->val != fast->val){
                return false;
            }
            slow = slow->next;
            fast = fast->next;
        }
        return true;
    }
    ListNode* reverse(ListNode* head){
        if(head == NULL){
            return NULL;
        }
        ListNode * pre, *cur, *nex;
        pre = NULL;
        cur = head;
        nex = head;
        while(cur != NULL){
            nex = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nex;
        } 
        return pre;
    }
};

栈

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        // if(head == NULL) return false;
        ListNode* cur = head;
        stack<int> stk;
        int length = 0;
        while(cur != NULL){
            stk.push(cur->val);
            cur = cur->next;
            length++;
        }
        int index = 0;
        while(head != NULL && index < length /2){
            if(head->val != stk.top()){
                return false;
            }
            head = head->next;
            stk.pop();
            index++;
        }
        return true;
    }
};

递归