[Leetcode] [Python]19:remove Nth Node from End of List

#-*-Coding:utf8-*-
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__author__ = ' [email protected] '

19:remove Nth Node from End of List
https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

Given A linked list, remove the nth node from the end of the 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 would always be valid.
Try to do the in one pass.

===comments by dabay===
The two-pointer-to-go solution should not conform to test instructions. A pass is required, you two hands at the same time, in fact, 2 times pass.
Idea One:
When traversing, put each node in a stack, and then more n pop to the corresponding location to delete the node. The space complexity is the length of the ListNode.
Idea two:
Use a queue of size n+1 to record the n nodes before the pointer. When the pointer is to the end, delete the second element in the queue. The complexity of space is n+1.
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# Definition for singly-linked list.
Class ListNode:
def __init__ (self, x):
Self.val = X
Self.next = None

Class Solution:
# @return A ListNode
def removenthfromend (self, head, N):
cursor = New_head = ListNode (0)
New_head.next = Head
Queue = []
While cursor:
Queue.append (cursor)
If Len (queue) > n + 1:
Queue.pop (0)
cursor = Cursor.next
Previous = Queue.pop (0)
To_del = Queue.pop (0)
Previous.next = To_del.next
Return New_head.next
# cursor = New_head = ListNode (0)
# New_head.next = Head
# stack = []
# while cursor:
# stack.append (cursor)
# cursor = Cursor.next
# while n > 1:
# Stack.pop ()
# n = n-1
# To_del = Stack.pop ()
# previous = Stack.pop ()
# Previous.next = To_del.next
# return New_head.next


def main ():
Sol = solution ()
root = ListNode (1)
N2 = ListNode (2)
N3 = ListNode (3)
N4 = ListNode (4)
Root.next = N2
N2.next = N3
N3.next = N4
Sol.removenthfromend (root, 1)
While Root:
Print "%s"% Root.val,
root = Root.next
Print "End"


if __name__ = = ' __main__ ':
Import time
Start = Time.clock ()
Main ()
Print "%s sec"% (Time.clock ()-start)

[Leetcode] [Python]19:remove Nth Node from End of List