[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：

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 };