[leetcode]Reverse Linked List II @ Python

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法 python

本文介绍了如何在链表中翻转从位置 m 到 n 的区间，并在原地完成操作，无需额外空间且只遍历一次链表。

原题地址：https://oj.leetcode.com/problems/reverse-linked-list-ii/

题意：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

解题思路：翻转链表的题目。

代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @param m, an integer
    # @param n, an integer
    # @return a ListNode
    def reverseBetween(self, head, m, n):
        if head == None or head.next == None:
            return head
        dummy = ListNode(0); dummy.next = head
        head1 = dummy
        for i in range(m - 1):
            head1 = head1.next
        p = head1.next
        for i in range(n - m):
            tmp = head1.next
            head1.next = p.next
            p.next = p.next.next
            head1.next.next = tmp
        return dummy.next