Remove Nth Node From End of List

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.
使用两个指针扫描，当第一个指针扫描到第N个结点后， 第二个指针从表头与第一个指针同时向后移动， 当第一个指针指向空节点时，另一个指针就指向倒数第n个结点了 

/**
 * 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* ans=new ListNode(0);
        ans->next=head;
        ListNode* temp=ans;
        for(int i=0;i<n;i++)
            head=head->next;
        while(head!=NULL){
            temp=temp->next;
            head=head->next;
        }
        temp->next=temp->next->next;
        return ans->next;
    }
};