从单链表中获取前n个值

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

def get_n_values_from_linked_list(head, n):
    count = 0
    new_head = ListNode()  # 创建新链表的头节点
    current = new_head  # 当前节点指针

    while head and count < n:
        current.next = ListNode(head.val)  # 将当前节点的值添加到新链表中
        current = current.next  # 移动当前节点指针
        head = head.next  # 移动原链表的指针
        count += 1

    return new_head.next  # 返回新链表的头节点的下一个节点，即为前n个节点的值

# 示例用法
# 创建一个单链表：1 -> 2 -> 3 -> 4 -> 5
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)

n = 3  # 获取前3个节点的值
result = get_n_values_from_linked_list(head, n)

# 输出结果
while result:
    print(result.val)
    result = result.next