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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 head head = first_start = ListNode(next=head) counter = 0 while head: if counter < left - 1: head = head.next elif counter == left - 1: first_end = head head = head.next elif counter == left: second_start = head pre = head head = head.next elif counter < right: tmp = head.next head.next = pre pre = head head = tmp elif counter == right: second_end = head third_start = head.next head.next = pre pre = None # 拼接 first_end.next = second_end second_start.next = third_start return first_start.next else: break counter += 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 = head pre = dummy_node for _ in range(left - 1): pre = pre.next cur = pre.next for _ in range(right - left): next = cur.next cur.next = next.next next.next = pre.next pre.next = next return 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。 