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：碰巧之前看了一篇相关的文章，思路就是可以采用两个指针来扫，第一个指针先走n步，然后一起走，当第一个指针指向NULL的时候，第二个指针所指就是要删除的节点，当删除的是头节点时候直接返回head->next即可。

代码如下，时间复杂度应该是O(n):

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