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.
# 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
        """
        def reverse(root, prep, k):
            cur = root
            pre = None
            next = None
            while cur and k > 0:
                next = cur.next
                cur.next = pre
                pre = cur
                cur = next
                k -= 1
            root.next = next
            prep.next = pre
            return pre
            
        dummy = ListNode(-1)
        dummy.next = head
        k = 1
        p = dummy
        start = None
        while p:
            if k == m:
                start = p
            if k == n + 1:
                reverse(start.next, start, n - m + 1)
                return dummy.next
            k += 1
            p = p.next