一、题目描述
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
示例：
输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL
二、代码实现
首先找到需要反转的链表部分进行反转，反转完毕后连接回去
# 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
"""
if head == None: return head
dummy = ListNode(-1)
dummy.next = head
start = dummy
for i in range(m-1):
start = start.next
end = cur = start.next
pre = None
for j in range(n-m+1):
last = cur.next
cur.next = pre
pre = cur
cur = last
start.next = pre
end.next = cur
return dummy.next
