234. Palindrome Linked List

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题解一：利用stack来判断回文，这里注意奇偶的处理，利用快慢指针找中点。

/**
 * 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 || head->next == NULL) return true;
        stack ST;
        ListNode * slow = head;
        ListNode * fast = head->next;
        while(fast && fast->next){
            ST.push(slow->val);
            cout<val<next->next;
            slow= slow->next;
        }
        if(fast){
            ST.push(slow->val);
        }
        slow = slow->next;
        while(slow){
            int x = ST.top();
            ST.pop();
            if(slow->val != x){
                return false;
            }
            slow = slow->next;
        }
        return true;
        
    }
};

题解二：
将后半段进行翻转，然后比较。

/**
 * 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 || head->next == NULL){
            return true;
        }
        ListNode * fast = head->next;
        ListNode * slow = head;
        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
        }
        if(fast){
            slow = slow->next;
        }
        ListNode * rhead = reverse(slow);
        ListNode * lcurr = head, * rcurr = rhead;
        while(rcurr){
            if(lcurr->val != rcurr->val){
                return false;
            }
            lcurr = lcurr->next;
            rcurr = rcurr->next;
        }
        return true;
        
    }
private:
    ListNode * reverse(ListNode *phead){
        ListNode * prev = NULL;
        ListNode * cur = phead;
        while(cur){
            ListNode *tmp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
    }
};