Remove Nth Node From End of List

Remove Nth Node From End of List

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.

用3个指针，一个指向list的头，用来返回，2个用来遍历，一前一后，长度为n，直到后一个指针为null，将前一个指针的下一个节点删除掉。判断是要分两种情况，一种是要删除头结点，一种是非头结点。

    ListNode * p, *q = new ListNode(0), *r;        q -> next = head;        p = head;        for(int i = 0; i < n; i++){  //将一个指针走n步            p = p -> next;        }        if(p == NULL){               //如果已经到了尾部，说明要删除头结点            return head -> next;        }        r = head;        while(p){            head = head -> next;            q = q -> next;            p = p -> next;        }        q -> next = head -> next;        delete head;        return r;    }