题目描述

Leetcode 24 题主要考察的链表的反转,而 25 题是 24 的拓展版,加上对递归的考察。

对题目做一下概述:

提供一个链表,给定一个正整数 k, 每 k 个节点一组进行翻转,最后返回翻转后的新链表。

k 的值小于或等于链表的长度,如果节点总数不是 k 的整数倍,将最后一组剩余的节点保持原有顺序。

注意:

  • 算法只能使用常数的空间
  • 不能单纯的改变节点内部的值,需要进行节点交换。

举例:

Example:
Given 1->2->3->4->5.
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5 Given 1->2.
For k = 2, you should return: 2->1
For k = 3, you should return: 1->2

解题思路

主要考察来了我们对链表的反转已经递归思想的熟悉度。

对链表反转时,可以借助三个指针,pre,current,next

  • pre 表示被翻转节点的前一个节点
  • current 表示正在被翻转的节点
  • next 表示正在被翻转的下一个节点

每次翻转时:

  • next 指向 current 的下一个
  • current 的 next 指向 prev
  • 同时 current 和 prev 向前一步

在每次翻转后:

  • current 表示下一个需要被翻转组中的第一个节点
  • prev 表示当前翻转后的新链表头
  • 原来的头 head 成了被翻转组中的尾节点

递归的处理

首先看到 K 个一组,想到到局部处理,并且局部处理翻转的方式是一致的,自然联想到递归。

递归的出口就是,被翻转后的新链表头或者未满足条件无需翻转的链表头。

代码实现:

# Question: Reverse Nodes in k-Group
# Given a linked list, reverse the nodes of a linked list k at a time and
# return its modified list.
#
# k is a positive integer and is less than or equal to the length of the linked
# list. If the number of nodes is not a multiple of k then left-out nodes in the
# end should remain as it is. # Example:
# Given 1->2->3->4->5.
# For k = 2, you should return: 2->1->4->3->5
# For k = 3, you should return: 3->2->1->4->5 # Note:
# Only constant extra memory is allowed.
# You may not alter the values in the list's nodes, only nodes itself
# may be changed. # Definition for singly-linked list. class ListNode:
def __init__(self, x):
self.val = x
self.next = None class Solution: def reverseKGroup(self, head: ListNode, k: int) -> ListNode: # check if head is NULL List
if head is None or head.next is None:
return head # get the number of nodes
list_length = 0
new_head = head
while new_head is not None:
list_length += 1
new_head = new_head.next # If the length of nodes is less than the number of group
if list_length < k:
return head # calculate the number of groups that can be reversed
number_of_groups = int(list_length / k) return self.swapPairs(head, k, number_of_groups) def swapPairs(self, head: ListNode, k, number_of_groups, number_of_reversed_groups=0) -> ListNode: prev = None
current = head
n = k # reverse the node due to the count of n
# After the reversal is completed,
# prev is the new head of the group that has been reversed.
# current points the head of the next group to be processed.
# head is the new end of the group that has been reversed.
while current and n > 0:
n -= 1
next = current.next
current.next = prev
prev = current
current = next # after a group of nodes is reversed, then increase 1.
number_of_reversed_groups += 1 # determine whether to reverse the next group
if current is not None and number_of_reversed_groups < number_of_groups:
head.next = self.swapPairs(
current, k, number_of_groups, number_of_reversed_groups)
else:
head.next = current
return prev def print_list_node(self, head: ListNode):
result = ''
while head is not None:
result += str(head.val) + '->'
head = head.next
print(result.rstrip('->')) if __name__ == '__main__':
l1 = ListNode(1)
l2 = ListNode(2)
l3 = ListNode(3)
l4 = ListNode(4)
l5 = ListNode(5) l1.next = l2
l2.next = l3
l3.next = l4
l4.next = l5 solution = Solution()
solution.print_list_node(l1)
reversed_l1 = solution.reverseKGroup(l1, 2)
solution.print_list_node(reversed_l1)

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