LeetCode 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:    ListNode *removeNthFromEnd(ListNode *head, int n) {        ListNode *dummy = new ListNode(-1);        dummy->next = head;        ListNode *pre = dummy;        ListNode *fast = head, *slow = head;        while(n--)            fast = fast->next;        //fast为空时，slow到达要删除节点        while(fast != NULL) {            fast = fast->next;            pre = slow;            slow = slow->next;        }        pre->next = slow->next;        head = dummy->next;        delete dummy;        return head;    }};