[leetcode每日一題]11.8

1684. 統計一緻字元串的數目

給你一個由不同字元組成的字元串 ​

​allowed​

​ 和一個字元串數組 ​

​words​

​ 。如果一個字元串的每一個字元都在 ​

​allowed​

​ 中，就稱這個字元串是 一緻字元串 。

請你傳回 ​

​words​

​ 數組中 一緻字元串 的數目。

示例 1：

輸入：allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
輸出：2
解釋：字元串 "aaab" 和 "baa" 都是一緻字元串，因為它們隻包含字元 'a' 和 'b' 。      

示例 2：

輸入：allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
輸出：7
解釋：所有字元串都是一緻的。      

示例 3：

輸入：allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
輸出：4
解釋：字元串 "cc"，"acd"，"ac" 和 "d" 是一緻字元串。      

• ​

  ​1 <= words.length <= 104​

  ​
• ​

  ​1 <= allowed.length <= 26​

  ​
• ​

  ​1 <= words[i].length <= 10​

  ​
• ​

  ​allowed​

  ​ 中的字元 互不相同 。
• ​

  ​words[i]​

  ​ 和 ​

  ​allowed​

  ​ 隻包含小寫英文字母。

Solution

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        cnt = len(words)
        for word in words:
            for letter in word:
                if letter not in allowed:
                    cnt -= 1
                    break
        return cnt      

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        mask = 0
        for c in allowed:
            mask |= 1 << (ord(c) - ord('a'))
        res = 0
        for word in words:
            mask1 = 0
            for c in word:
                mask1 |= 1 << (ord(c) - ord('a'))
            res += (mask1 | mask) == mask
        return res

作者：力扣官方題解
連結：https://leetcode.cn/problems/count-the-number-of-consistent-strings/solutions/1953831/tong-ji-yi-zhi-zi-fu-chuan-de-shu-mu-by-38356/
來源：力扣（LeetCode）
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