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.

题意转换一下就是求解单链表的倒数第k个节点，解法是设置两个指针，一个在前一个在后，之间的距离为k，同时向前移动，有点类似滑动窗口，当第一个指针到达链表尾的时候，第二个指针即为倒数第k个节点，具体的见编程之美

ListNode *removeNthFromEnd(ListNode *head, int n){    if(head == NULL) return NULL;    ListNode *q = head,*p = head;    for(int i = 0 ; i < n - 1; ++ i)        q = q -> next;    ListNode *pPre = NULL;    while(q->next){        pPre = p;        p = p->next;        q = q->next;    }    if(pPre == NULL){        pPre = p->next;        delete p;    }else{        pPre ->next = p ->next;        delete p;    }    return head;}