原題連結：
https://leetcode.com/problems/reverse-linked-list-ii/description/

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

下面是第一遍寫的演算法，醜得一B，邊際條件太多，應該可以寫出更加簡介的程式碼，wr了好多次，終於A了，弱雞，其實這道題分別用兩個指標移動到m和n的位置，就轉化為

class Solution:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """
        #1->2->3->4->5->NULL
        #1->4->3->2->5->NULL
        #first用來儲存1的指標位置
 

        #second用來儲存2的指標位置
        #former與latter用來前後滑動，進行
        first = head
        if m == n:
            return head
        if m == 1:
            first = head
            time = n-m-1
        elif m == 2:
            first = head
            time = n-m
        else:
            for i in range
 
 (m-2):
                first = first.next
            time = n-m
        second = first.next
        former = second
        current = second.next
        if current:
            latter = second.next.next
        else:
            latter = None
        if m == 1 and n == 2:
            second.next = first
            first.next = current
            return second
        for j in range (time):
            current.next = former
            former = current
            current = latter
            if current:
                latter = latter.next
            else:
                latter = None
            #if not (latter or current):
                #break
        if m == 1:
            second.next = first
            first.next = current
            return former
        else:
            first.next = former
            second.next = current
            return head