actually i stuck in this prog, not gettting it exactly, i want user to input natural no. of 4 digit and i want its all possible permutations of the number.. plz help me..

10 Years
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Last Post by anifeelings
public class Permutations {

    // print N! permutation of the characters of the string s (in order)
    public  static void perm1(String s) { perm1("", s); }
    private static void perm1(String prefix, String s) {
        int N = s.length();
        if (N == 0) System.out.println(prefix);
        else {
            for (int i = 0; i < N; i++)
               perm1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));


    // print N! permutation of the elements of array a (not in order)
    public static void perm2(String s) {
       int N = s.length();
       char[] a = new char[N];
       for (int i = 0; i < N; i++)
           a[i] = s.charAt(i);
       perm2(a, N);

    private static void perm2(char[] a, int n) {
        if (n == 1) {
        for (int i = 0; i < n; i++) {
            swap(a, i, n-1);
            perm2(a, n-1);
            swap(a, i, n-1);

    // swap the characters at indices i and j
    private static void swap(char[] a, int i, int j) {
        char c;
        c = a[i]; a[i] = a[j]; a[j] = c;

    public static void main(String[] args) {
       int N = Integer.parseInt(args[0]);
       String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
       String elements = alphabet.substring(0, N);

wilol this work out..?? please do reply..

Edited by Nick Evan: Added code-tags

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