写个最小编辑距离吧~

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