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：第一種，把連結串列Reverse

/**
 * 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) 
    {
        int lenth=0;
        int nth=0;
        ListNode dummyNode(0);
        dummyNode.next=head;
        ListNode *p=&dummyNode;
        while(p->next)
        {
            lenth++;
            p=p->next;
        }
        p=&dummyNode;
        nth=lenth-n;
        while(nth--&&p->next)
        {
            p=p->next;
        }
        p->next=p->next->next;
        return dummyNode.next;
    }
};