想法：用 hash set紀錄已經出現過的字串，嘗試所有字串組合並檢查是否已經出現過，未出現過則加入字串

最後輸出 hash set 大小即可

Time Complexity : O(n!) for there are O(n!) possible strings

Space Complexity : O(n!) for the hash set and the recursion

class Solution {

public:

unordered_set<string> appear ;

void constructTiles (vector<int>& frequency , string now) {

for(int i = 0 ; i < 26 ; i++) {

if ( frequency[i] > 0 ) {

frequency[i]-- ;

for(int insertindex = 0 ; insertindex <= now.size();insertindex++) {

string newstr ;

if (insertindex > 0)

newstr += now.substr(0 , insertindex) ;

newstr += ('A' + i) ;

if (insertindex < now.size())

newstr += now.substr(insertindex) ;

if ( appear.find(newstr) == appear.end() ) {

appear.insert(newstr) ;

constructTiles(frequency , newstr) ;

}

}

frequency[i]++ ;

}

}

}

int numTilePossibilities(string tiles) {

vector<int> frequency(26) ;

for(auto &c : tiles)

frequency[ c - 'A']++ ;

constructTiles(frequency , "") ;

return appear.size() ;

}

};

// 法二：只需考慮剩餘字母即可，用遞迴決定第 i 個字母填什麼，接著考慮 i + 1 ... n

回傳剩餘字母可組成的種類數量 + 單獨該字母組成的一種字串

Time Complexity : O(n!) for the recursion

Space Complexity : O(n!) for the recursion

class Solution {

public:

int constructTiles (vector<int>& frequency , int size ) {

int count = 0 ;

for(int i = 0 ; i < 26 ; i++) {

if ( frequency[i] > 0 ) {

frequency[i]-- ;

count += constructTiles(frequency , size + 1) + 1 ;

frequency[i]++ ;

}

}

return count ;

}

int numTilePossibilities(string tiles) {

vector<int> frequency(26) ;

for(auto &c : tiles)

frequency[ c - 'A']++ ;

return constructTiles(frequency , 0) ;

}

};