我有一個輸入字典 -dict_input目標為keys，源為values。一個目的地可以有一個或多個來源。dict_input = {'C411':['C052'],'C052':['C001','C002'], 'C001':['9001'], 'C002':['9002']}在上面dict_input，終端目的地是，C411而初始來源是9001和9002。我正在嘗試為終端目的地提出源路徑C411。預期輸出形式為list-[['C411', 'C052', 'C001', '9001'], ['C411', 'C052','C002', '9002']]我有這個代碼：def get_source(node, dict_input, source=[]): if node in dict_input: source.append(node) for i in dict_input[node]: if i != node: get_source(i, dict_input, source) else: source.append(node) else: source.append(node) return source return sourcedict_input = {'C052':['C001','C002'], 'C411':['C052'], 'C001':['9001'], 'C002':['9002']} print(get_source('C411', dict_input, []))輸出是合并成一個列表的兩個源路徑 -['C411', 'C052', 'C001', '9001', 'C002', '9002']如何修改我的代碼以獲取每個源路徑的單獨列表？
1 回答
精慕HU
TA貢獻1845條經驗 獲得超8個贊
pop完成訪問節點后不要忘記從當前路徑
如果遇到“葉子”（不是鍵的節點 ID），則在輸出列表中存儲當前路徑的副本
警惕損壞的數據，例如循環鏈接 - 將有助于保留set訪問過的節點
上面的示例實現：
def get_source(root, dict_input):
# output list
path_list = []
# current path
cur_path = []
# visited set
visited = set()
# internal recursive helper function
def helper(node):
cur_path.append(node)
# normal node
if node in dict_input:
if not node in visited:
visited.add(node)
for child in dict_input[node]:
helper(child)
# else: cycle detected, raise an exception?
# leaf node
else:
# important: must be a copy of the current path
path_list.append(list(cur_path))
cur_path.pop()
# call this helper function on the root node
helper(root)
return path_list
測試：
>>> dict_input = {'C411':['C052'],'C052':['C001','C002'], 'C001':['9001'], 'C002':['9002']}
>>> get_source('C411', dict_input)
[['C411', 'C052', 'C001', '9001'], ['C411', 'C052', 'C002', '9002']]
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