1 题目
链表翻转的进阶版，即只翻转链表的某一部分。这就要求我们找到待翻转链表的头结点和尾节点，然后再翻转一下就可以了。
2 思想
- 翻转位置left 到 位置right 的链表节点
- 本任务拆解成两部分
- 确定pre_head, tail_next 2. 链表翻转
3 代码
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
pre_head = None
tmp = head
# step 1. 确定pre_head
cnt = 1
while(cnt < left):
pre_head = tmp
tmp = tmp.next
cnt += 1
cnt = 1
tail_next = head
while(cnt < right):
tail_next = tail_next.next
cnt +=1
tail_next = tail_next.next
if pre_head is None:
start = head # 从第一个头结点开始翻转
else:
start = pre_head.next
# step 2. 链表翻转
nxt = start.next
tmp = start
while(start and nxt != tail_next):
tmp2 = nxt.next #
nxt.next = start
start = nxt
nxt = tmp2
if pre_head is None:
head = start
else:
pre_head.next = start
tmp.next = tail_next
return head
