Leetcode 19 删除链表的倒数第 N 个结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

找到链表的倒数第 n 个节点，然后删除该节点即可。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        count = 0
        current = head
        while current:
            count += 1
            current = current.next
        pos = count - n
        index = 0
        current = head

        if pos == 0:
            return head.next

        while current:
            index += 1
            if index == pos and current.next is not None:
                current.next = current.next.next
                break
            else:
                current = current.next
        return head