[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;

     }

 };

巴特西

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

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