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.

思路：主要考虑删除头节点的情况，因此最好加一个哨兵进行处理。

/**
 * 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) {
        if(head == NULL)    return NULL;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        
        ListNode *fast = dummy;
        ListNode *slow = dummy;
        while(n-- > 0)
            fast = fast->next;
        
        while(fast->next != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }
        
        ListNode *delNode = slow->next;
        slow->next = delNode->next;
        delete delNode;
        return dummy->next;
    }
};