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

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.

Follow up:

Could you do this in one pass?

解題思路：一遍遍歷法，雙指標法，先讓一個之前前進n個，再讓兩個同時遍歷，直到第一個指標的next為NULL

/**
 * 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(n==0)
            return head;
        ListNode *p=head,*q=head;
        for(int i=0;i<n;i++)
            q=q->next;
        while(q!=NULL&&q->next!=NULL){
            p=p->next;
            q=q->next;
        }
        if(q==NULL)
            return head->next;
        else{
            p->next=p->next->next;
            return head;
        }
    }
};