Reverse Linked List II.md

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

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL

My code:

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        if head==None or head.next==None:
            return head
        origin_head=head
        count=1
        list_value=[]
        while head!=None:
            if count>=m and count<=n:
                list_value.append(head.val)
                if count==m:
                    head1=head
            
            count+=1
            head=head.next
        list_value=list_value[::-1]
        count=m
        while count<=n:
            head1.val=list_value[count-m]
            head1=head1.next
            count+=1
        return origin_head
        

Solution: https://leetcode.com/problems/reverse-linked-list-ii/