211. String permutation "Lintcode by Java"

Description

Given-strings, write a method to decide if one was a permutation of the other.

Example

abcdis a permutation bcad of, but are not abbe a permutation ofabe

Problem solving: A similar problem has been encountered, the simpler way is to convert the string into a character array, and then sort, compare the number of each bit is equal. This is less efficient and the code is as follows:

1  Public classSolution {2     /**3      * @parama:a String4      * @paramb:a String5      * @return: A Boolean6      */7      Public Booleanpermutation (string A, String B) {8         //Write your code here9         if(A.length ()! =b.length ())Ten             return false; One         Char[]a =A.tochararray (); A         Char[]B =B.tochararray (); - Arrays.sort (a); - Arrays.sort (b); the          for(inti = 0; i < a.length; i++){ -             if(A[i]! =B[i]) -                return false; -         } +         return true; -     } +}

It can also be done through a hash table: if the number of each element is equal, and the total number is the same, the string is exactly equal, refer to: 51212982.

211. String permutation "Lintcode by Java"