Remove Nth Node From End of List

来源：Leetcode

原帖：http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

题目：
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.
Note:
Given n will always be valid. Try to do this in one pass.
Solution: head---back------front------>NULL
                   |          |
                   ---> n <----

代码：

 /**  
  * 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) {  
         // corner case: 头结点,使用dummy node  
         ListNode dummy(0);  
         ListNode* back = &dummy;  
         ListNode* front = &dummy;      
         dummy.next = head;  
         while (n--) {  
             front = front->next;  
         }  
         while (front->next) {  
             front = front->next;  
             back = back->next;  
         }  
         ListNode *node = back->next;  
         back->next = node->next;  
         delete node;  
         return dummy.next;  
     }  
 };