(Medium #19) Remove Nth Node From End of List

Problem

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

My solution

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

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        out = None
        
        vallist = []
        while True:            
            vallist.append(head.val)
            head = head.next
            if head == None:
                break
        
        outlist = vallist[0:len(vallist)-n] + vallist[len(vallist)-n+1:]        
        for i in range(len(outlist)-1,-1,-1):
            temp = ListNode(outlist[i])
            temp.next = out
            out = temp
        return out            

Runtime: 32ms, Memory: 14MB

참고 (입력 만들기)

x_list=[1,2,3,4,5]
x = None
pre = ListNode(0)
for i in range(len(x_list)):
    pre = ListNode(x_list[i])
    pre.next = x
    print(id(x), id(pre))
    x = pre
    print(id(x), id(pre))