【LeetCode】C++ :简单题 - 字符串 1684. 统计一致字符串的数目

1684. 统计一致字符串的数目

难度简单5

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。

1、哈希表 

    class Solution {
    public:
        int countConsistentStrings(string allowed, vector<string>& words) {
            unordered_set<char> ust;
            for(auto ch: allowed){
                ust.insert(ch);
            }
            int res = 0;
            for(int i = 0; i < words.size(); i++){
                int j = 0;
                for(j; j < words[i].size(); j++){
                    char ch = words[i][j];
                    if(!ust.count(ch)){
                       break;
                    }
                }
                if(j == words[i].size()){
                    res ++;
                }
            }
            return res;
        }
    };

 

class Solution {
public:
    int countConsistentStrings(string allowed, vector<string>& words) {
        int res = 0;
        int f[26] = {0};
        int fwords[26] = {0};

        for(int i = 0; i < allowed.size(); i++){
            f[allowed[i]-'a']++;
        }
        for(int i = 0; i < words.size(); i++){
            int fwords[26] = {0};
            for(int j = 0; j < words[i].size(); j++){
                fwords[words[i][j]-'a']++;
            }
            res++;
            for(int k = 0; k < 26; k++){
                if(fwords[k] && f[k] == 0){
                    res--;
                    break;
                }
            }
        }    
        return res;
        
    }
};

 

 

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