Hunter的大杂烩 技术学习笔记

测试
a=("ab/cd/ef", "ab/cd/ef/g", "ab/cd/")
输出
{'ab/cd/': 1}

[code]

#in_set 应是一个有序的字符串集合
def remove_duplicate_string(in_set):
    if len(in_set) < 2:
        return in_set

    duplicate_str_dict = {}
    same_str_dict = {}

    old_str = ""
    same_str = ""
    for str in in_set:
        if len(same_str) > 1:
            if str.find(same_str) >= 0: # same_str 应该是两个字符串的交集，如果能跟same_str匹配，则直接pass
                continue
        if len(old_str) > 0:
            same_str = get_same_string(old_str, str)
            print(f"{same_str} vs {old_str} vs {str}")
        if len(same_str) > 1:
            same_str_dict[same_str] = 1
        old_str = str
    if len(same_str_dict) < 1 or len(same_str_dict) == len(in_set):
        return in_set
    if len(same_str_dict) < len(in_set):
        return remove_duplicate_string(same_str_dict)
    return same_str_dict

[/code]