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

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* removeNthFromEnd(ListNode* head, int n) {
12         if(!head) return head;
13         
14         ListNode* first = head; 
15         for(int i = 0 ; i < n - 1; first = first->next, ++i);
16         
17         if(!first->next){
18             first = head;
19             head = head->next;
20             delete first;
21             return head;
22         }
23         
24         ListNode* second= head;
25         for(;first->next->next; first = first->next, second = second->next);
26         
27         ListNode* next = second->next; 
28         second->next = next->next; 
29         delete next;
30         
31         return head;
32     }
33 };