题目：

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.

代码：

从链表中删除从右向左数第N个元素，况且只能遍历链表一遍。

所以，需要定义两个指针，相差是N个元素，同时向前遍历，当靠前的元素走到最后的时候，在后面的元素刚好与最后一个元素距离为N，删除即可。

于是：

#Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution(object):
    
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        p = head
        q = head
        if head.next ==None:return []
        for i in range(0,n):
            q = q.next
        if not q : return head.next
        while q.next:
            p = p.next
            q = q.next
        temp = p.next.next
        p.next = temp
        return head
        
if __name__==‘__main__‘:
    arr = [1,2]
    p = ListNode(arr[0])
    head = p
    for i in arr[1:]:
        p.next = ListNode(i)
        p = p.next
    s=Solution()
    q=s.removeNthFromEnd(head,2)
    while q:
        print (q.val)
        q = q.next

19. Remove Nth Node From End of List