Remove Nth Node From End of List

题目描述：

Given a linked list, remove the n^th 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