题解 | 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
        dummy = ListNode(-101)
        dummy.next = head
        pre, cur = dummy, dummy
        for _ in range(n):#使得pre，cur相隔n个节点
            cur = cur.next
        while cur.next:#到达要删除节点的前一个节点
            cur = cur.next
            pre = pre.next
        pre.next = pre.next.next
        return dummy.next