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;
}
};