leetcode 19. Remove Nth Node From End of List

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：
最直接的思路，应该是先统计链表节点的个数num，然后再找到第num-n-1那个节点x，把那个x节点的next指向x->next->next就行了。但是这个很明显不能在一遍遍历解决。
可以用双指针的思想初始p=head q=head。开始走n个节点到p,然后从p->next开始，和q一起走，知道p->next为nullptr,那么此时的q就是上个思路的x节点。那么q->next=q->next->next就行了。
代码如下：

 * 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) {
        if(head == NULL || head->next == NULL)
            return NULL;
        ListNode *p = head;
        ListNode *q = head;
        while(--n >= 0) p = p->next;
        if(p == NULL) return head->next;
        
        p = p->next;
        while(p != NULL) {
            p = p->next;
            q = q->next;
        }
        q->next = q->next->next;
        return head;
    }
};