单链列表的定义。

Question

我有这样一个ListNode

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

和一个链表

 Input: l1: 1->2->4, ll2:1->3->4

如何将数字添加到列表中？

我尝试了

head = ListNode(0)
node1 = ListNode(head)
node1.next = node2 #but node2 has not been declared
node2 = ListNode(2)
node2.next = node3 #node3 has not been declared
node3 = ListNode(4)
node3.next = null

我认为我需要一些d = defaultdict(int).之类的Listnode东西，然后可以在分配之前使用d []。

Answer 1

我可以通过在节点类中添加工厂方法来解决此问题-

单链列表的定义。

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
    def link(value):
        node = ListNode(value)
        self.next = node
        return node

然后您的初始化代码将变为...

head = ListNode(0)
head.link(2).link(4).link(6)

Answer 2

这将运行：

head = ListNode(0)
node1 = ListNode(head) # <-- is this really what you want
node2 = ListNode(2)
node3 = ListNode(4)

node1.next = node2
node2.next = node3
node3.next = null # <-- or maybe it won't

但这不是您想要的吗

head = ListNode(0)
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(4)

head.next = node1
node1.next = node2
node2.next = node3
# node3.next = None  # <-- this is redundant

如果您这样定义班级：

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x, next=None):
        self.val = x
        self.next = next

您可以这样做：

head = ListNode(0, ListNode(1, ListNode(2, ListNode(4))))

但是也许您想要所有对节点的中间引用

将元素写入链表

2 个答案:

单链列表的定义。