题目链接
难度：中等 类型： 链表
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
解题思路
0.保存头节点
1.找到反转部分的前一个节点，保存为start
2.翻转第m到n位链表，记录第m个节点为node_m，第n个节点为node_n，第n+1个节点为end
3.连接链表，start.next = node_n, node_m.next = end
4.返回头节点
代码实现
# 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 or not head.next or m==n:
return head
dummy = ListNode(0)
dummy.next = head
start = dummy
for i in range(m-1):
start = start.next
end = cur = start.next
pre = None
for i in range(n-m+1):
next = cur.next
cur.next = pre
pre = cur
cur = next
start.next = pre
end.next = cur
return dummy.next
