[Leetcode] remove nth node from the end of list 删除链表倒数第n各节点

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.

这题比较简单，使用快慢指针，找到倒数第n个结点的前驱即可，然后连接其前驱和后继即可，值得注意的是，两个while的条件之间的关系。代码如下：

 /**
  * 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)
     {
         ListNode *nList=new ListNode(-);
         nList->next=head;
         ListNode *pre=nList;
         ListNode *fast=head;
         ListNode *slow=head;
         int num=;

         while(num++<n)
         {
             fast=fast->next;
         }
         while(fast->next)
         {
             pre=pre->next;
             slow=slow->next;
             fast=fast->next;
         }
         pre->next=slow->next;

         return nList->next;
     }
 };