Print out all permutations of an Array [duplicate]

Answer 1

The trick is to return a special value (false in the code below) from nextPerm when it was the last permutation (i.e. when array become sorted in descending order):

import java.util.*;

public class Main {
    public static boolean nextPerm(List a) {
        int i = a.size() - 2;
        while (i >= 0 && a.get(i) >= a.get(i + 1))
            i--;

        if (i < 0)
            return false;

        int j = a.size() - 1;
        while (a.get(i) >= a.get(j))
            j--;

        Collections.swap(a, i, j);
        Collections.reverse(a.subList(i + 1, a.size()));
        return true;
    }
    ...

Then you can use the loop (note that the array required be sorted in ascending order initially):

    ...
    public static void main(String[] args) {
        List a = Arrays.asList(new Integer[] {1, 2, 3, 4});
        do {
            System.out.println(a);
        } while (nextPerm(a));
    }
}

You can try this code here: http://ideone.com/URDFsc