03 2023 档案

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

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