You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed.
Return the number of consistent strings in the array words.
Example 1
Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
Output: 2
Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.
Example 2
Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
Output: 7
Explanation: All strings are consistent.
Example 3
Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
Output: 4
Explanation: Strings "cc", "acd", "ac", and "d" are consistent.
Constraints
- 1 <= words.length <= 1e4
- 1 <= allowed.length <= 26
- 1 <= words[i].length <= 10
- The characters in allowed are distinct.
- words[i] and allowed contain only lowercase English letters.
Solution
自己最先想到的是哈希表，因为只有26个小写字母
class Solution {
public:
int countConsistentStrings(string allowed, vector<string>& words) {
vector<int> hash(26, 0);
for (int i = 0; i < allowed.size(); i++) {
hash[allowed[i] - 'a'] = 1;
}
int ans = 0;
for (auto& s : words) {
int flag = 1;
for (int i = 0; i < s.size(); i++) {
if (!hash[s[i] - 'a']) flag = 0;
}
if (flag) ans++;
}
return ans;
}
};
题解用的是位运算表示集合，十分高级
class Solution {
public:
int countConsistentStrings(string allowed, vector<string>& words) {
int mask = 0;
for (auto c : allowed) {
mask |= 1 << (c - 'a');
}
int res = 0;
for (auto &&word : words) {
int mask1 = 0;
for (auto c : word) {
mask1 |= 1 << (c - 'a');
}
if ((mask1 | mask) == mask) {
res++;
}
}
return res;
}
};
