题目：
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.
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