題目

給你一個由不同字符組成的字符串 allowed 和一個字符串?dāng)?shù)組 words 。如果一個字符串的每一個字符都在 allowed 中，就稱這個字符串是 一致字符串 。

請你返回 words 數(shù)組中 一致字符串 的數(shù)目。

示例 1：

輸入：allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
輸出：2
解釋：字符串 "aaab" 和 "baa" 都是一致字符串，因為它們只包含字符 'a' 和 'b' 。
示例 2：

輸入：allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
輸出：7
解釋：所有字符串都是一致的。
示例 3：

輸入：allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
輸出：4
解釋：字符串 "cc"，"acd"，"ac" 和 "d" 是一致字符串。

提示：

1 <= words.length <= 104
1 <= allowed.length <= 26
1 <= words[i].length <= 10
allowed 中的字符 互不相同 。
words[i] 和 allowed 只包含小寫英文字母。

解題思路

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        res = 0
        ## 通過Set方式來進行判斷抽取，復(fù)雜度降低
        allowedSet = set(allowed)
        for word in words:
            wordSet = set(word)
            if wordSet.issubset(allowedSet):
                res += 1
            # res += 1
            # for i in wordSet:
            #     if i not in allowedSet:
            #         res -= 1
            #         break
        return res

if __name__ == '__main__':
    # allowed = "ab"
    # words = ["ad","bd","aaab","baa","badab"]
    allowed = "abc"
    words = ["a", "b", "c", "ab", "ac", "bc", "abc"]
    ret = Solution().countConsistentStrings(allowed, words)
    print(ret)