Description:

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.

解法：

由于只允许一次遍历，所以我们不能用一次完整的遍历来统计链表中元素的个数，而是遍历到对应位置就应该移除了。那么我们需要用两个指针来帮助我们解题，pre和cur指针。首先cur指针先向前走N步，如果此时cur指向空，说明N为链表的长度，则需要移除的为首元素，那么此时我们返回head->next即可，如果cur存在，我们再继续往下走，此时pre指针也跟着走，直到cur为最后一个元素时停止，此时pre指向要移除元素的前一个元素，我们再修改指针跳过需要移除的元素即可。代码如下：

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        p = head;
        tmp = p;
        for i in range(n):
            q = tmp.next;
            tmp = q;
        if(q == None):
            return head.next;
        while(q.next != None):
            p = p.next;
            q = q.next;
        tmp = p.next;
        if(tmp != None):
            p.next = tmp.next;
        return head;