[leetcode][list][two pointers] Remove Nth Node From End of List

题目：

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.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public://注意空指针、可能删除head所指节点    ListNode* removeNthFromEnd(ListNode* head, int n) {if (NULL == head || n <= 0) return head;ListNode *ahead = head, *behind = head, *pre = NULL;//ahead指向最后一个节点的时候behind指向要删除的节点，pre指向要删除节点的前驱int i = 1;while (i < n && ahead){++i;ahead = ahead->next;}if (!ahead) return head;//n大于链表长度while (ahead && ahead->next){ahead = ahead->next;pre = behind;behind = behind->next;}if (!pre){ListNode *p = head;head = head->next;delete p;p = NULL;}else{pre->next = behind->next;delete behind;behind = NULL;}return head;}};