LeetCode 026. 重排链表

And_Ii2026-05-21 10:00

LCR 026. 重排链表

给定一个单链表 L的头节点 head ，单链表 L 表示为：

L0 → L1 → ... → Ln-1 → Ln

请将其重新排列后变为：

L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → ...

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1：

复制代码

输入: head = [1,2,3,4]
输出: [1,4,2,3]

示例 2：

复制代码

输入: head = [1,2,3,4,5]
输出: [1,5,2,4,3]

提示：

链表的长度范围为 [1, 5 * 104]

1 <= node.val <= 1000

Definition for singly-linked list.

class ListNode:

def init(self, val=0, next=None):

self.val = val

self.next = next

class Solution:
def reverse(self,head1):
if head1 is None:
return None
pre=None
cur=head1
while cur is not None:
cur_new=cur.next
cur.next=pre
pre=cur
cur=cur_new
return pre

复制代码

  def reorderList(self, head: ListNode) -> None:
      """
      Do not return anything, modify head in-place instead.
      """
      if head is None:
          return
      
      slow=head
      fast=head
      while fast.next is not None and fast.next.next is not None:
          slow=slow.next
          fast=fast.next.next
      
      half_2=slow.next
      slow.next=None
      tail=self.reverse(half_2)
      
      cur_h=head
      while tail is not None:
          cur_h_new=cur_h.next
          cur_h.next=tail
          tail=tail.next
          cur_h.next.next=cur_h_new
          cur_h=cur_h_new