反转链表（LeetCode）

好好学习Py2024-08-08 12:23

题目

给你单链表的头节点，请你反转链表，并返回反转后的链表

解题

python 复制代码

class ListNode:
    def __init__(self, value=0, next=None):
        self.value = value
        self.next = next


def reverse_linked_list_recursive(head: ListNode) -> ListNode:
    # 空链表或单节点链表
    if not head or not head.next:
        return head

    # 递归反转子链表
    new_head = reverse_linked_list_recursive(head.next)

    # 处理当前节点
    head.next.next = head
    head.next = None

    return new_head


# 辅助函数：创建链表
def create_linked_list(elements):
    if not elements:
        return None
    return ListNode(elements[0], create_linked_list(elements[1:]))


# 辅助函数：打印链表
def print_linked_list(head: ListNode):
    current = head
    while current:
        print(current.value, end=" -> " if current.next else "\n")
        current = current.next


# 测试代码
if __name__ == '__main__':
    # 创建链表: 1 -> 2 -> 3 -> 4 -> 5
    elements = [1, 2, 3, 4, 5]
    head = create_linked_list(elements)

    print("原始链表：")
    print_linked_list(head)

    reversed_head = reverse_linked_list_recursive(head)

    print("反转后的链表：")
    print_linked_list(reversed_head)

原始链表：

1 -> 2 -> 3 -> 4 -> 5

反转后的链表：

5 -> 4 -> 3 -> 2 -> 1