写个最小编辑距离吧~

    技术2025-09-05  48

    def recursive_edit_distance(str1, str2, cost): if not str1: return len(str2) * cost["add"] if not str2: return len(str1) * cost["delete"] if str1[-1] == str2[-1]: return recursive_edit_distance(str1[:-1], str2[:-1], cost) else: return min(recursive_edit_distance(str1[:-1], str2, cost) + cost["delete"], recursive_edit_distance(str1, str2[:-1], cost) + cost["add"], recursive_edit_distance(str1[:-1], str2[:-1], cost) + min(cost["replace"], cost["delete"] + cost["add"])) def dp_edit_distance(str1, str2, cost): value = [[0] * (len(str2) + 1) for i in range(len(str1) + 1)] # 这里不能使用value = [[0] * (len(str2) + 1)] * (len(str1) + 1)进行初始化,因为所有行都是第一行的引用, # 它们是同步变化的. 最好全部用range方式初始化 for i in range(len(str1) + 1): value[i][0] = i * cost["delete"] for j in range(len(str2) + 1): value[0][j] = j * cost["add"] for i in range(1, len(str1) + 1): for j in range(1, len(str2) + 1): if str1[i - 1] == str2[j - 1]: value[i][j] = value[i - 1][j - 1] else: value[i][j] = min(value[i - 1][j] + cost["delete"], value[i][j - 1] + cost["add"], value[i - 1][j - 1] + min(cost["replace"], cost["delete"] + cost["add"])) for i in range(len(value)): print(value[i]) return value[-1][-1] str_ = ["", "a", "abc", "efa", "bcc"] cost = {"delete": 1, "add": 1, "replace": 1} # 如果不同的操作对应的花费不同的话,可能出现的一个情况是,使用一次删除和一次插入代替一次替换更加划算 for i in range(len(str_)): for j in range(i + 1, len(str_)): print("str_[{}], str_[{}]:".format(i, j), recursive_edit_distance(str_[i], str_[j], cost) == dp_edit_distance(str_[i], str_[j], cost)) str1 = "acef" str2 = "abce" print(dp_edit_distance(str1, str2, cost)) # TODO: 打印最小编辑路径
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