92_reverse_linked_list_II.py

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
        dummy = ListNode(0, head)
        count = 1
        prev = dummy
        while (prev != None) and (count < left):
            count += 1
            prev = prev.next
        curr = prev.next
        for i in range(left, right):
            next_node = curr.next
            curr.next = next_node.next
            next_node.next = prev.next
            prev.next = next_node
        return dummy.next

'''
    This is my personal record of solving Leetcode Problems. 
    If you have any questions, please discuss them in [Issues](https://github.com/mengxinayan/leetcode/issues).
    Copyright (C) 2020-2022  mengxinayan

    This program is free software: you can redistribute it and/or modify
    it under the terms of the GNU General Public License as published by
    the Free Software Foundation, either version 3 of the License, or
    (at your option) any later version.

    This program is distributed in the hope that it will be useful,
    but WITHOUT ANY WARRANTY; without even the implied warranty of
    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
    GNU General Public License for more details.

    You should have received a copy of the GNU General Public License
    along with this program.  If not, see <http://www.gnu.org/licenses/>.
'''