比标准答案更好的解 Py3 | #删除链表的倒数第n个节点#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param n int整型 
# @return ListNode类
#
class Solution:
    def removeNthFromEnd(self , head: ListNode, n: int) -> ListNode:
        # write code here
        if not head:
            return head

        ptr_r = head
        ptr_l = head
        ptr_l_prev = None
        ptr_l_step = n
        
        # find
        while ptr_r:
            ptr_r = ptr_r.next
            if ptr_l_step > 0:
                ptr_l_step -= 1
            else:
                ptr_l_prev = ptr_l
                ptr_l = ptr_l.next

        target = ptr_l
        target_prev = ptr_l_prev

        # delete
        if target_prev: # is midd or end
            target_prev.next = target.next
        else: # is start
            head = target.next

        print(target_prev.val if target_prev else "nil", target.val)
        return head