[Leetcode]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.

删除链表的倒数第n个元素。n保证合法。要求走一趟搞定。

代码来自这位大神

简直牛X！

ListNode *removeNthFromEnd(ListNode *head, int n){ListNode dummy(-1);dummy.next = head;ListNode *pre = &dummy;while (--n > 0) head = head->next;while (head->next){pre = pre->next;head = head->next;}pre->next = pre->next->next;return dummy.next;}