LeetCode:Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

思路：先从头遍历等到长度length，在从头遍历，根据当前结点次序与长度length和n的关系确定要删除的结点。

 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         ListNode *p=head,*q;
13         int length=0;
14         while(p!=NULL)
15         {
16             length++;
17             p=p->next;
18         }
19         p=head;
20         head=q;
21         q->next=p;
22         while(p!=NULL)
23         {
24             if(length==n)
25             {
26                q->next=p->next;
27                break;
28             }
29             else
30             {
31                 q=p;
32                 p=p->next;
33                 length--;
34             }
35         }
36         return head->next;
37     }
38 };