Given a linked list, remove the n th 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.

Note:
Given n will always be valid.
Try to do this in one pass.

這題比較簡單，使用快慢指針，找到倒數第n個結點的前驅即可，然後連接其前驅和後繼即可，值得註意的是，兩個while的條件之間的關系。代碼如下：

 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     {
13         ListNode *nList=new ListNode(-1);
14         nList->next=head;
15         ListNode *pre=nList;
16         ListNode *fast=head;
17         ListNode *slow=head;
18         int num=0;
19 
20         while(num++<n)
21         {
22             fast=fast->next;
23         }   
 
24         while(fast->next)
25         {
26             pre=pre->next;
27             slow=slow->next;
28             fast=fast->next;
29         } 
30         pre->next=slow->next;
31 
32         return nList->next;
33     }
34 };

[Leetcode] remove nth node from the end of list 刪除鏈表倒數第n各節點