题解 | #判断一个链表是否为回文结构#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 the head
     * @return bool布尔型
     */
    ListNode* middleNode(ListNode* head)
    {
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast && fast->next)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }

    ListNode* reverList(ListNode* head)
    {
        if(head->next == nullptr)
        {
            return head;
        }

        ListNode* n1 = nullptr;
        ListNode* n2 = head;
        ListNode* n3 = n2->next;
        while(n2)
        {
            n2->next = n1;
            n1 = n2;
            n2 = n3;
            if(n3)
            {
                n3 = n3->next;
            }
        }

        return n1;
    }

    bool isPail(ListNode* head) 
    {
        ListNode* mid = middleNode(head);
        ListNode* newlist = reverList(mid);
        while(head && newlist)
        {
            if(head->val != newlist->val)
            {
                return false;
            }
            head = head->next;
            newlist = newlist->next;
        }
        return true;
    }
};