[LeetCode]Remove Nth Node From End of List

又一道链表题。CareerCup上也有类似题目，trick和“部分链表逆序”一样，由于不知道链表的长度，second指针比first指针先走n步，然后在同时向前移动，直到second指针到达链表尾。注意删除头节点的情况。

假设链表长度为N，倒数第k个节点是正数第N-k个节点(k<=N)，前驱指针p指向第N-k-1个节点，后向指针q指向第N个节点。

Given a linked list, remove the nth node from the end of list and return its head.

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.

代码

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL)
            return head;
            
        ListNode *p = NULL;
        ListNode *q = head;
        for(int i=1; i<n; ++i)
            q = q->next;
        while(q->next!=NULL)
        {
            if(p)
                p = p->next;
            else
                p = head;
            q = q->next;
        }
        if(p)
        {
            ListNode *tmp = p->next;
            p->next = tmp->next;
            delete tmp;
        }
        else
        {
            ListNode *tmp = head;
            head = head->next;
            delete tmp;
        }
        
        return head;
    }
};