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。
class Solution { public: int countConsistentStrings(string allowed, vector<string>& words) { int count = words.size(); for(int i = 0; i < words.size(); ++ i) { for(int j = 0;j < words[i].size(); ++ j) { if(allowed.find(words[i][j]) == -1) { count--; break; } } } return count; } };
class Solution {public: int countConsistentStrings(string allowed, vector<string>& words) { int count = words.size(); for(int i = 0; i < words.size(); ++ i) { for(int j = 0;j < words[i].size(); ++ j) { if(allowed.find(words[i][j]) == -1) { count--; break; } } } return count; }};
作者：Hello World链接：https://leetcode.cn/problems/count-the-number-of-consistent-strings/solutions/1885586/class-solution-public-int-countconsisten-8344/来源：力扣（LeetCode）著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。