题目描述：
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.
好久没做题，少了股锐气。这题思路才是关键。
solution：
ListNode *removeNthFromEnd(ListNode *head, int n) { if(head == NULL) return NULL; ListNode dummyHead(0); dummyHead.next = head; ListNode *fast = &dummyHead, *slow = &dummyHead; while(n--) fast = fast->next; while(fast->next != NULL) { fast = fast->next; slow = slow->next; } slow->next = slow->next->next; return dummyHead.next; }
原文链接：https://leetcode.com/discuss/1656/is-there-a-solution-with-one-pass-and-o-1-space
Remove Nth Node From End of List
原文：http://www.cnblogs.com/gattaca/p/4328915.html