[leetcode] 19 delete the penultimate node of the linked list

I'll give you a list , Delete the last of the linked list n Nodes , And return the head node of the list .

Example 1：

Input ： head = [1,2,3,4,5], n = 2
Output ： [1,2,3,5]

Example 2：

Input ： head = [1], n = 1
Output ： []

Example 3：

Input ： head = [1,2], n = 1
Output ： [1]

Tips ：

• The number of nodes in the list is sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

Advanced ： Can you try a scan implementation ？

# The linked list is stored in the sequence for random access 

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        arr = []
        length = 0
        new = ListNode(0,head)# Construct a linked list of leading nodes 
        tmp = new
        while tmp:
            arr.append(tmp)
            tmp = tmp.next
            length += 1
        arr[length-n-1].next = arr[length-n-1].next.next
        return new.next

# Push 

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        arr = []

        new = ListNode(0,head)
        tmp = new
        while tmp:
            arr.append(tmp)
            tmp = tmp.next
        
        for i in range(n):
            arr.pop()
        pre = arr.pop()
        pre.next = pre.next.next
        return new.next

# Double pointer . Advanced ： A scan 
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        new = ListNode(0,head)
        first = new
        second = head# namely new.next
        for i in range(n):
            second = second.next
        
        while second:
            first = first.next
            second = second.next
            
        first.next = first.next.next
        return new.next