题解 | #调整牛群顺序#

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【1】题目考察的知识点：链表、快慢指针

【2】题目解答方法的文字分析 题目有误，应为倒数第n个节点

1. 定义快慢指针：快慢指针为虚拟头节点，快指针先走n步，此时快慢同时走，快到终点时，慢的下一个节点为所要移动节点
2. 进行链表段的拼接组合

【3】本题解析所用的编程语言：python

【4】完整且正确的编程代码

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def moveNthToEnd(self, head: ListNode, n: int) -> ListNode:
        if head is None or head.next is None:return head
        if n==1:return head
        dummy=ListNode(-1)
        dummy.next=head
        fast=slow=dummy
       
        for i in range(n):
            if fast is not None:fast=fast.next
            else: return head
        while fast.next:
            fast=fast.next
            slow= slow.next
        tail=slow.next
        slow.next= slow.next.next
        fast.next=tail
        tail.next=None
        return dummy.next