Are two string Permutation - Java Algorithm

Java examples for Algorithm:String

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Description

Are two string Permutation

Demo Code

import java.util.HashMap;

public class Main {
  public static void main(String[] args){
    String s_1 = "ABCD";
    String s_2 = "DBCA";
    String s_3 = "XYZ";
    String s_4 = "WXYZ";
    String s_5 = "";
    System.out.println("isPermutation_1 Method"); 
    System.out.println("'ABCD' and 'DBCA' are permutations?: " + isPermutation_1(s_1, s_2));
    System.out.println("'ABCD' and 'XYZ' are permutations?: " + isPermutation_1(s_1, s_3));
    System.out.println("'ABCD' and 'WXYZ' are permutations?: " + isPermutation_1(s_1, s_4));
    System.out.println("'ABCD' and '' are permutations?: " + isPermutation_1(s_1, s_5));
    //from  ww  w . j a va2  s  .c  o  m
    System.out.println("isPermutation_2 Method"); 
    System.out.println("'ABCD' and 'DBCA' are permutations?: " + isPermutation_2(s_1, s_2));
    System.out.println("'ABCD' and 'XYZ' are permutations?: " + isPermutation_2(s_1, s_3));
    System.out.println("'ABCD' and 'WXYZ' are permutations?: " + isPermutation_2(s_1, s_4));
    System.out.println("'ABCD' and '' are permutations?: " + isPermutation_2(s_1, s_5));
  }
  public static boolean isPermutation_1(String str_1, String str_2){ // With the assumption that all characters are unique
    if(str_1.length() != str_2.length()) return false;
    HashMap<Character, Boolean> map = new HashMap<Character, Boolean>();
    for(int i=0; i<str_1.length(); i++){
      if(map.get(str_1.charAt(i)) == null)
        map.put(str_1.charAt(i), true);
    }
    
    for(int j=0; j<str_2.length(); j++){
      if(map.get(str_2.charAt(j)) == null) return false; 
    }
    return true;
  }
  
  public static boolean isPermutation_2(String str_1, String str_2){ //Not all characters are unique, and use recursion to list out all permutations ans stor in a set
    return false;
  }
}

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