Python 算法解析：用优美代码轻松掌握 Leetcode 61 - 旋转链表

2024-02-16 03:23:46

算法概述

旋转链表 是一个经典的链表操作问题，它要求你将一个链表中的元素向右移动 k 个位置。例如，如果链表为 1->2->3->4->5，并且 k = 2，那么旋转后的链表应为 4->5->1->2->3。

解决 leetcode 61 旋转链表问题的基本思路是：

遍历链表，找到倒数第 k 个节点。
将链表从倒数第 k 个节点处断开，形成两个子链表。
将第二个子链表连接到第一个子链表的末尾，形成一个新的链表。
返回新的链表。

Python 代码

def rotate_list(head, k):
  """
  Rotates a linked list by k positions.

  Args:
    head: The head of the linked list.
    k: The number of positions to rotate the list.

  Returns:
    The head of the rotated linked list.
  """

  # Find the length of the linked list.
  length = 0
  current = head
  while current:
    length += 1
    current = current.next

  # If k is greater than the length of the linked list,
  # then we need to rotate the list multiple times.
  k %= length

  # If k is 0, then the list does not need to be rotated.
  if k == 0:
    return head

  # Find the new head of the linked list.
  current = head
  for _ in range(k):
    current = current.next

  # Break the linked list at the new head.
  new_head = current
  current = head
  while current.next != new_head:
    current = current.next
  current.next = None

  # Connect the second half of the linked list to the first half.
  current = new_head
  while current.next:
    current = current.next
  current.next = head

  # Return the new head of the linked list.
  return new_head