删除链表的倒数第N个节点

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {

        assert(n >= 0);
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode* p = dummyHead;
        ListNode* q = dummyHead;
        for (int i = 0; i < n + 1; i++)
        {
            assert(q);
            q = q -> next;
        }
        
        while ( q != NULL)
        {
            p = p -> next;
            q = q -> next;
        }
        ListNode* delNode = p -> next;
        p -> next = delNode -> next;
        delete delNode;

        ListNode* retNode = dummyHead->next;
        delete dummyHead;

        return retNode;
    }
};

输入: 1->2
输出: false

输入: 1->2->2->1
输出: true

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode* slowp = head;
        ListNode* fastp = head;
        ListNode* revp = NULL;
        while (fastp && fastp->next) {
            fastp = fastp->next->next;
            ListNode* tmp = slowp->next;
            slowp->next = revp;
            revp = slowp;
            slowp = tmp;
        }
        if (fastp) 
            slowp = slowp->next;
        while ( slowp && revp) {
            if (slowp->val != revp->val) return false;
            slowp = slowp->next;
            revp = revp->next;
        }
        return true;
    }
};