String permutation using for loops

Question

I have to print all the possible permutations of the given input string.

Using the code below I get aaaa bbb ccc now in next iteration I want to print aaa aab aac. aba aca and so on. Please guide me about it.

String s = "abc";
char ch;
ArrayList<Character> input = new ArrayList<Character>();

public static void main (String [] args)
{
  String  s= "abc";
  int count ; 
  char ch;
  ArrayList<Character> input = new ArrayList<Character>();
  for (int i=0; i < s.length(); i++)
  {
    ch = s.charAt(i);
    input.add(ch);         
  }       


  for (int i=0; i <= input.size(); i++)
  {
    for(int j=0; j < input.size(); j++)
    {
      System.out.print(input.get(i));
    } 
    System.out.println();
  } 
}

Why are you using an ArrayList<Character> to store String characters while every character can be read using s.charAt(i)? — BackSlash
– BackSlash, Commented May 1, 2013 at 9:14
@BackSlash i have placed specific string here , but input is not specific that can be abc123 etc . — Sikander
– Sikander, Commented May 1, 2013 at 9:15
Sounds like you want every possible permutation of the input characters. — Duncan Jones
– Duncan Jones, Commented May 1, 2013 at 9:33
possible duplicate of Generating all permutations of a given string — Duncan Jones
– Duncan Jones, Commented May 1, 2013 at 9:44

Spinning · Accepted Answer · 2013-05-01 09:42:56Z

3

You can use recursive function. Example

private static String text = "abcd";

public static void main(String[] args) {
    loopPattern(text, 0, "");
}

private static void loopPattern(String source, int index, String res) {
    if (source == null || source.length() == 0) {
    return;
    }
    if (index == source.length()) {
        System.out.println(res);
        return;
    }
    for (int i = 0; i < source.length(); i++) {
        loopPattern(source, index + 1, res + source.charAt(i));
    }
}

answered May 1, 2013 at 9:42

Spinning

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Comments

StreakyCobra · Accepted Answer · 2013-05-01 09:43:22Z

2

A recursive version which is not dependent of the number of characters:

class Test
{
    public static void main(String[] args)
    {
        if(args.length != 1)
        {
            System.out.println("Usage: java Test <string>");
            System.exit(1);
        }

        String input = args[0];

        iterate(input, "", input.length());
    }

    public static void iterate(String input, String current, int level)
    {
        if(level == 0)
        {
            System.out.println(current);
            return;
        }

        for(int i=0; i < input.length(); i++)
        {
            iterate(input, current + input.charAt(i), level-1);
        }
    }

answered May 1, 2013 at 9:43

StreakyCobra

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Comments

SeniorJD · Accepted Answer · 2013-05-01 09:27:17Z

1

In current implementation, you should use:

for (int i=0; i <= input.size(); i++) {
   for(int j=0; j < input.size(); j++) {
      for(int k=0; k < input.size(); k++) {
         System.out.print(input.get(i));
         System.out.print(input.get(j));
         System.out.print(input.get(k));
         System.out.println();
      }
   }
}

But IMHO it is better to use s.charAt(i) instead of input.get(i).

answered May 1, 2013 at 9:27

SeniorJD

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3 Comments

Duncan Jones Over a year ago

It sounds like the OP wants every possible permutation of the input characters. In which case, this solution doesn't scale because it requires one for loop per character.

Sikander Over a year ago

this is very much related to what i was told in class , i just did not try it same time and totally lost the idea . this is really helpful

Abhishek Sengupta Over a year ago

dude this is not the correct solution

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