Brother word Query

One word, one word, and one letter exchange, can get another word, such as army-> mary, to become a brother word. Provide a word and find its brother in the dictionary. Describes the data structure and query process. Analysis: There are many solutions to this problem on the Internet. I used the Count sorting method to solve this problem. The ASCII value of a letter in any word (which is assumed to be in lowercase) ranges from 97 to (97 + 26 ). The English letters of sibling words are the same, but the order is different. We set two arrays, for example, A [26] and B [26], to traverse the letters in A word. The letter ASCII code of the word is set to variable I, the ASCII code of the letter of the brother word is set to variable j, then, A [i-97] ++, B [j-97] ++. Finally traverse the letter of the word, the traversal variable is set to I, if for all I, A [i-97] = B [i-97], then the two words are brother words. Through this method, we can obtain an algorithm with the complexity of O (n. Code:

Bool IsBrotherWords (const std: string & str1, const std: string & str2) {int nLength1 = str1.length (); int nlengh2 = str2.length (); // if (nLength1! = Nleng2) return false; // set two Arrays: int a [26], B [26]; // initialize for (int I = 0; I <26; ++ I) {a [I] = B [I] = 0 ;}// start counting the letters in the word for (int I = 0; I <nLength1; ++ I) {a [str1.at (I)-97] ++; B [str2.at (I)-97] ++;} // checks whether the count is the same, if not, this is not the sibling word for (int I = 0; I <nLength1; I ++) {if (a [str1.at (I)-97]! = B [str1.at (I)-97]) return false;} return true ;}

 

The code above can be optimized. An array is used to add 1 to each letter count of a word, and 1 to each letter count of a sibling word. If the count of the last array is 0, the two words are brother words.

Bool IsBrotherWords2 (const std: string & str1, const std: string & str2) {int nLength1 = str1.length (); int nlengh2 = str2.length (); // if (nLength1! = Nleng2) return false; // design number array int a [26]; for (int I = 0; I <26; ++ I) {a [I] = 0 ;}// start counting for (int I = 0; I <nLength1; ++ I) {a [str1.at (I) -97] ++; a [str2.at (I)-97] --;} // if the last count is not 0, it is not a sibling word for (int I = 0; I <nLength1; I ++) {if (a [str1.at (I)-97]! = 0) return false;} return true ;}

 

The algorithm time complexity is O (2 * Length + 26), and Length is the word Length. Test code:

int _tmain(int argc, _TCHAR* argv[])  {        std::string str1("army");      std::string str2("mary");    bool b= IsBrotherWords(str1,str2);  bool b2=IsBrotherWords2(str1,str2);         return 0;  }