LeetCode 19. 删除链表的倒数第 N 个结点

思路：
先翻转链表再删除第n个节点再翻转链表
删除第n个节点需要pre指向head节点再用cur记录删除节点的前一个节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* head) {
        ListNode* res = new ListNode(0);

        while (head) {
            ListNode* temp = head ->next;
            head ->next = res ->next;
            res ->next = head;

            head = temp;
        }

        return res ->next;
    }


    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* head_rev = reverse(head);

        ListNode* pre = new ListNode(0);
        pre ->next = head_rev;
        ListNode* cur = pre;

        n = n -1;
        while (n --) cur = cur ->next;

        cur->next = cur ->next ->next;

        return reverse(pre ->next);

    }
};