摘要:
def get_permute2(nums): ret = [] path = [] def backtracking(nums,usage_list): if len(path) == len(nums): ret.append(path[:]) return for i in range(0,l 阅读全文
posted @ 2023-03-26 17:47
fufu1
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摘要:
#回溯算法 获取字典所有值的键的集合def get_dict_path(data): ret = [] path =[] def traceback(path,data): if len(path)>0 and not isinstance(data, dict): ret.append(path[ 阅读全文
posted @ 2023-03-25 17:19
fufu1
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