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LeetCode 92. 反转链表 II

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。
示例 1：
输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
示例 2：
输入：head = [5], left = 1, right = 1
输出：[5]
提示：
链表中节点数目为 n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n
进阶： 你可以使用一趟扫描完成反转吗？

# 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: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:if left == right:return headhead = first_start = ListNode(next=head)counter = 0while head:if counter < left - 1:head = head.nextelif counter == left - 1:first_end = headhead = head.nextelif counter == left:second_start = headpre = headhead = head.nextelif counter < right:tmp = head.nexthead.next = prepre = headhead = tmpelif counter == right:second_end = headthird_start = head.nexthead.next = prepre = None# 拼接first_end.next = second_endsecond_start.next = third_startreturn first_start.nextelse:breakcounter += 1

时间复杂度 O(n)：一个大循环最多遍历链表完整一次，计O(n)。共O(n)。
空间复杂度 O(1)：常量。共 O(1)。

还是官解写的简洁

class Solution:def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:# 设置 dummyNode 是这一类问题的一般做法dummy_node = ListNode(-1)dummy_node.next = headpre = dummy_nodefor _ in range(left - 1):pre = pre.nextcur = pre.nextfor _ in range(right - left):next = cur.nextcur.next = next.nextnext.next = pre.nextpre.next = nextreturn dummy_node.next# 作者：力扣官方题解
# 链接：https://leetcode.cn/problems/reverse-linked-list-ii/solutions/634701/fan-zhuan-lian-biao-ii-by-leetcode-solut-teyq/
# 来源：力扣（LeetCode）
# 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。