[leetcode]92. 反转链表 II

1.题目：

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
说明: 1 ≤ m ≤ n ≤ 链表长度。

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

2.代码：

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

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        if not head:
            return None
		# cur 指向第一个要转的位置，pre指向cur之前
        cur, prev = head, None
        while m > 1:
            prev = cur
            cur = cur.next
            m, n = m - 1, n - 1
		'''
		tail 一直指向3,因为3就是最后一个
		prev 一直指向
		1->2—>{3-> 4-> 5}->6->7 ==> 1->2—>{5-> 4-> 3}->6->7
		   |   |   |                   |   |       |    |
		  con cur third               con pre     tail cur
		  pre                      
		'''
        tail, con = cur, prev
        while n:
            third = cur.next
            cur.next = prev
            prev = cur
            cur = third
            n -= 1
        # con == None 即prev == None，即从第1个开始置换，那么head = prev
        if con:
            con.next = prev
        else:
            head = prev
        tail.next = cur
        return head