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anuj
permutstion of numbers in java
2 Answer(s)      2 years and 6 months ago
Posted in : Java Beginners


Is it possible to enter the number in a char so that permutation of a number can be acheived which will be in the form of string????? here is the coding i did...it worked really well when i initialized the char as(1,2,3).but its not working when i want the input from user.. .like if user enters 3...then string shud bcum 123 and permutation shud be obtained...plz help me i need to submit coding tomorrow itself...... import java.util.*; public class major1

{ int n=0; char c[]; public major1() {

        System.out.println("Enter the number of processors=");
            Scanner input=new Scanner(System.in);
        n=input.nextInt(); 
        for(int j=1;j<=n;j++)
        {
            c[n]=j;
        }
        String s=new String(c);
        showPattern("", s);
}






        public void showPattern(String st, String s) 
        {
            if (s.length() <= 1)

            System.out.println("\n"+st + s);
            else
            {
                for (int i = 0; i < s.length(); i++) 
                {
                    try 
                    {
                        String newString = s.substring(0, i)
                        +s.substring(i + 1);
                        showPattern(st + s.charAt(i), newString);
                    }   
                    catch (Exception e) 
                    {
                    e.printStackTrace();
                    }
                }
            }
        }
        public static void main(String args[])
        {
            major1 m1=new major1();
        }

}


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View Answers

April 12, 2012 at 5:26 PM


import java.util.*;

public class PermutationExample extends ArrayList<String> {

public static void main(String[] args) {

int[] set = {1, 2, 3};
PermutationExample retVal = permutations(set, 3);
Collections.sort(retVal);
for (int k = 0; k < retVal.size(); k++) {
System.out.printf(" %s\n", retVal.get(k));
}
}
         private static PermutationExample permutations(int[] set, int choices) {
            PermutationExample perm =new PermutationExample();
            String resetStr, setStr;
            resetStr = setStr = "";
            for (int j = 0; j < choices; j++) {
            resetStr += String.format("%d", set[0]);
            setStr = resetStr;
            }
            java.util.Random rand = new java.util.Random();
            int max = 1;
            int k = 1;
            int nextDown = set.length;
            while (k <= choices) {
            max *= nextDown;
            nextDown--;
            k++;
            }
            while (perm.size()< max) {
            setStr = "";
            java.util.ArrayList<Integer> ele =new java.util.ArrayList<Integer>();
            while (ele.size() < choices) {
            int anInt = set[rand.nextInt(set.length)];
            if (!ele.contains(anInt)) {
            ele.add(anInt);
            }
            }
            setStr =String.format("%d%d%d",ele.get(0), ele.get(1), ele.get(2));
            if (!perm.contains(setStr)) {
            perm.add(setStr);
            }
            }
            return perm;
         }
        private boolean contains(String str) {
            return ((ArrayList) this).contains(str);
        }
      }


April 12, 2012 at 5:27 PM


import java.util.*;

public class PermutationExample {
    public static void main(String args[]) throws Exception {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter String: ");
        String chars = input.next();
        showPattern("", chars);
    }

    public static void showPattern(String st, String chars) {
        if (chars.length() <= 1)
            System.out.println(st + chars);
        else
            for (int i = 0; i < chars.length(); i++) {
                try {
                    String newString = chars.substring(0, i)
                            + chars.substring(i + 1);
                    showPattern(st + chars.charAt(i), newString);
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
    }
}


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