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。

Remove Nth Node From End of List -- leetcode