题目
给你一个由不同字符组成的字符串 allowed 和一个字符串数组 words 。如果一个字符串的每一个字符都在 allowed 中，就称这个字符串是 一致字符串 。
请你返回 words 数组中 一致字符串 的数目。
示例 1： 输入：allowed = "ab", words = ["ad","bd","aaab","baa","badab"] 输出：2 解释：字符串 "aaab" 和 "baa" 都是一致字符串，因为它们只包含字符 'a' 和 'b' 。 示例 2： 输入：allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"] 输出：7 解释：所有字符串都是一致的。 示例 3： 输入：allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"] 输出：4 解释：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 只包含小写英文字母。
解题思路
class Solution: def countConsistentStrings(self, allowed: str, words: List[str]) -> int: res = 0 ## 通过Set方式来进行判断抽取，复杂度降低 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)
