Leetcode 19 Remove Nth Node From End of List 链表

删除从后往前数的第n个节点

我的做法是将两个指针first，second

先将first指向第n-1个，然后让first和second一起指向他们的next，直到first->next->next为空

最后只要删除second->next

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* removeNthFromEnd(ListNode* head, int n) {
         if(!head) return head;

         ListNode* first = head;
         for(int i =  ; i < n - ; first = first->next, ++i);

         if(!first->next){
             first = head;
             head = head->next;
             delete first;
             return head;
         }

         ListNode* second= head;
         for(;first->next->next; first = first->next, second = second->next);

         ListNode* next = second->next;
         second->next = next->next;
         delete next;

         return head;
     }
 };