LeetCode 92. 反转链表 II

予早2024-07-20 23:48

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？

python 复制代码

# 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: 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

python 复制代码

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。