92. 反转链表 II--PYTHON_92 反转链表2 迭代法python-优快云博客

本文链接：https://blog.youkuaiyun.com/hot7732788/article/details/103616516

本文介绍了一种在单链表中反转指定区间的算法实现。通过将链表转换为列表进行反转，再重新构建链表，实现了m到n位置节点的反转。文章详细展示了算法步骤，包括节点的遍历、列表的反转及链表的重构。

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自己：

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

class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """

        data_list = list()
        p = head

        while p:
            data_list.append(int(p.val))
            p = p.next
        data_f = data_list[:m-1]
        data_t = data_list[n:]
        data_m = data_list[m-1:n]
        data_m = data_m[::-1]
        data = data_f + data_m + data_t
        data = data[::-1]

        res = ListNode(-1)
        for i in data:
            tmp = ListNode(i)
            tmp.next = res.next
            res.next = tmp
        
        return res.next

大佬1：看的我头都晕了，有空在加上注释

在这里插入图片描述

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

class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """

        if not head:
            return None

        cur, prev = head, None
        while m > 1:
            prev = cur
            cur = cur.next
            m, n = m - 1, n - 1

        tail, con = cur, prev

        while n:
            third = cur.next
            cur.next = prev
            prev = cur
            cur = third
            n -= 1

        if con:
            con.next = prev
        else:
            head = prev
        tail.next = cur
        return head