题解 | #链表的回文结构#中间节点的查找+链表逆转

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    bool chkPalindrome(ListNode* phead) {
        
        if(!phead || !(phead->next))
        {
            return true;
        }

        ListNode* fast, * slow;//用于查找中间节点
        ListNode* n1, * n2, * n3;//用于逆转前半部分的链表

        fast = slow = phead;
        n1 = nullptr;
        n2 = phead;
        n3 = n2->next;

        //寻找中间节点的同时逆转前半部分的链表
        //每一次循环结束n2与slow指向同一个节点
        while(fast && fast->next)
        {
            //迭代查找中间节点
            slow = slow->next;
            fast = fast->next->next;

            //扭转指针指向
            n2->next = n1;
            //迭代
            n1 = n2;
            n2 = n3;
            n3 = n3->next;
        }

        //判断是否为回文串
        while(slow)
        {
            if(fast)
            {
                slow = slow->next;
                fast = fast->next;
            }
            if(n1->val != slow->val)
            {
                return false;
            }

            n1 = n1->next;
            slow = slow->next;
        }
        
        return true;
    }
};