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) {}
 * };
 */
#include <type_traits>
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
     */
    bool isPail(ListNode* head) {
        // write code here
        if(head == nullptr){
            return true;
        }
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast != nullptr && fast -> next != nullptr){
            fast = fast -> next -> next;
            slow = slow -> next;
        }
        slow = reverse(slow);
        while(slow != nullptr){
            if(head -> val != slow -> val){
                return false;
            }
            head = head -> next;
            slow = slow -> next;
        }
        return true;
    }

    //反转链表指针
    ListNode* reverse(ListNode* head){
        ListNode* prev = nullptr;
        while(head != nullptr){
            ListNode* next = head -> next;
            head -> next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
};