LeetCode 19. Remove Nth Node From End of List

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.
题意：
删除已链表尾部开始的第n个节点

题解：

清楚python的链表结构就行，比较简单

# 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
"""
i=1
res=[]
temp=head
while head.next != None:
head=head.next
i=i+1
j=1
while j<=i:
if j==i-n+1:
temp=temp.next
j=j+1
else:
res.append(temp.val)
temp=temp.next
j=j+1
return res