使用條件節點鏈接迭代鏈表

Question

我對編碼的想法很陌生，所以我不能 100% 確定我在問什么。 我有一個對象列表，其中包含一個框的 ID、可選選項的數量以及與其鏈接的框的 ID。

我正在嘗試在這些框中按順序迭代添加所有可以選擇的選項

例如，“Connect 6”總共有 3 個，“contect 7”總共有 2 個，“Connect 4”總共有 7 個

我有 2 個類 1 設置列表中的所有鏈接，然后應該遍歷它們的 1 個類。 我還沒有能夠得到總和部分，因為我所做的任何事情都會永遠打印出拳頭盒 ID，直到它崩潰 VS（不知道如何在它開始運行后阻止它做我的蠢事，即使關閉終端也不會）幫助）

class Node:
    def __init__(self, box_id):
        self.box_id = box_id
        self.options_number = None
        self.pre = None
        self.next = None
    def __repr__(self):
        return self.box_id

class LinkedList():
    def __init__(self, nodes=None):
        self.head = None
        self.reg_num = 0
        if nodes is not None:
            for node in nodes:
                box_id = node.box_id
                # print(f'One: {box_id}')
                # box_id = node.next
                # print(f'Two: {box_id}')
                # if box_id == []:
                #     print("I should really stop")
                while box_id != []:
                    print(box_id)
                    box_id = node.next

這是打印出節點的樣子。

nodes = node_setup(test)
for node in nodes:
   print(f'Question Text: {node.name}  Current: {node.box_id} Options: {node.options_number} Next Link: {node.next}')

Question Text: Connect 1  Current: {30706B51-789C-4AFA-998D-759297B95F99} Options: 1 Next Link: ['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
Question Text: Connect 2  Current: {EA65B6DA-90FE-412A-AF1F-83F994CCB5FA} Options: 3 Next Link: ['{77AC8029-95D8-4DA8-9B62-D487DBA742D6}']
Question Text: Connect 3  Current: {77AC8029-95D8-4DA8-9B62-D487DBA742D6} Options: 1 Next Link: ['{DF56E1B2-DD19-485D-B720-A5A28AA6C156}']
Question Text: Connect 5  Current: {B4045D8D-6E54-4E43-B820-4C70036F08AF} Options: 1 Next Link: ['{D1720CEC-81FF-4460-A5CB-86BC9B5AB7F5}']
Question Text: Connect 4  Current: {DF56E1B2-DD19-485D-B720-A5A28AA6C156} Options: 2 Next Link: ['{B4045D8D-6E54-4E43-B820-4C70036F08AF}', '{8C51E3B4-7302-4453-AEE0-7C9E659CA2B0}']
Question Text: Connect 6  Current: {8C51E3B4-7302-4453-AEE0-7C9E659CA2B0} Options: 2 Next Link: ['{D837ECDC-897D-4A91-89E3-F6222FE12F06}']
Question Text: Connect 7  Current: {D1720CEC-81FF-4460-A5CB-86BC9B5AB7F5} Options: 1 Next Link: ['{D837ECDC-897D-4A91-89E3-F6222FE12F06}']
Question Text: Connect 8  Current: {D837ECDC-897D-4A91-89E3-F6222FE12F06} Options: 1 Next Link: []

在這里它正在運行 LinkedList

for node in nodes:
    print(f'Question Text: {node.name}  Current: {node.box_id} Options: {node.options_number} Next Link: {node.next}')
plz_Work = LinkedList(nodes)

Question Text: Connect 1  Current: {30706B51-789C-4AFA-998D-759297B95F99} Options: 1 Next Link: ['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
Question Text: Connect 2  Current: {EA65B6DA-90FE-412A-AF1F-83F994CCB5FA} Options: 3 Next Link: ['{77AC8029-95D8-4DA8-9B62-D487DBA742D6}']
Question Text: Connect 3  Current: {77AC8029-95D8-4DA8-9B62-D487DBA742D6} Options: 1 Next Link: ['{DF56E1B2-DD19-485D-B720-A5A28AA6C156}']
Question Text: Connect 5  Current: {B4045D8D-6E54-4E43-B820-4C70036F08AF} Options: 1 Next Link: ['{D1720CEC-81FF-4460-A5CB-86BC9B5AB7F5}']
Question Text: Connect 4  Current: {DF56E1B2-DD19-485D-B720-A5A28AA6C156} Options: 2 Next Link: ['{B4045D8D-6E54-4E43-B820-4C70036F08AF}', '{8C51E3B4-7302-4453-AEE0-7C9E659CA2B0}']
Question Text: Connect 6  Current: {8C51E3B4-7302-4453-AEE0-7C9E659CA2B0} Options: 2 Next Link: ['{D837ECDC-897D-4A91-89E3-F6222FE12F06}']
Question Text: Connect 7  Current: {D1720CEC-81FF-4460-A5CB-86BC9B5AB7F5} Options: 1 Next Link: ['{D837ECDC-897D-4A91-89E3-F6222FE12F06}']
Question Text: Connect 8  Current: {D837ECDC-897D-4A91-89E3-F6222FE12F06} Options: 1 Next Link: []
['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
['{EA65B6DA-90FE-412A-AF1F-83F994CCB5FA}']
ect....

老實說，我對如何解決這個問題感到非常迷茫，我將不勝感激任何人都可以提供的幫助

Answer 1

我現在已經解決了。 我在錯誤的地方使用了 for。

它仍然不完美，因為我無法讓它在有 2 個鏈接的地方工作，但現在這讓我超越了永遠循環，能夠解決其余的問題

class LinkedList():
def __init__(self, nodes=None):
    self.head = None
    self.reg_num = 0
    if nodes is not None:
        box_id = nodes[0].box_id
        while box_id != []:
            node_range = len(nodes)
            for node in range(node_range):
                if box_id == nodes[node].box_id:
                    name = nodes[node].name
                    print(f'name {name}')
                    next_id = nodes[node].next
                    if next_id != []:
                        next_id = next_id[0] #This needs to be something more clever
                    box_id = next_id

和版畫

name Connect 1
name Connect 2
name Connect 3
name Connect 4
name Connect 5
name Connect 7
name Connect 8

================================================== =

編輯：現在弄清楚如何做整件事

start_id = nodes[0].box_id
plz_Work = LinkedList(nodes, start_id)

班級

class LinkedList():
def __init__(self, nodes, start_id):
    if nodes is not None:
        box_id = start_id
        while box_id != []:
            node_range = len(nodes)
            for node in range(node_range):
                if box_id == nodes[node].box_id:
                    name = nodes[node].name
                    print(f'name: {name}')
                    next_id = nodes[node].next
                    if next_id != []:
                        next_range = len(next_id)
                        if next_range < 2:
                            next_id = next_id[0]
                        else:
                            for next_num in range(next_range):
                                LinkedList(nodes, next_id[next_num])
                            next_id = []
                    box_id = next_id

打印

name: Connect 1
name: Connect 2
name: Connect 3
name: Connect 4
name: Connect 5
name: Connect 7
name: Connect 8
name: Connect 6
name: Connect 8