Remove Nth Node From End of List -- leetcode

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.

/** * 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) {        ListNode fake(0);        fake.next = head;        ListNode *p = &fake;        while (n--) p = p->next;        ListNode *q = &fake;        while (p->next) {                p = p->next;                q = q->next;        }        p = q->next;        q->next = p->next;        delete p;        return fake.next;    }};

用一对指针；再引入一个假头，以统一处理节点删除操作。

以上算法是基于条件：Given n will always be valid.，即假定n输入都是符合预期的。

否则，还应该加上适当的边界处理。