题解 | #判断一个链表是否为回文结构#咱们先找mid再reverse mid之后的链表，咱们拒绝一刀切

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 the head
     * @return bool布尔型
     */
    ListNode* getMid(ListNode* head) {
        if (!head) {
            return head;
        }
        ListNode *slow = head, *fast = head;
        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
    
    ListNode* reverse(ListNode* head) {
        if (!head) {
            return head;
        }
       ListNode *pre = nullptr, *curr = head;
        while(curr) {
            ListNode *next = curr->next;
            curr->next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
    bool isPail(ListNode* head) {
        // reverse mid 之后的链表，开始逐个递归。
        ListNode *mid = getMid(head);
        ListNode *l1 = head, *l2 = mid->next;
        mid->next = nullptr;
        l2 = reverse(l2);
        while(l1 && l2) {
            if (l1->val != l2->val) {
                return false;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
        return true;
        // write code here
    }
};