Generating all permutations of a given string

Question

What is an elegant way to find all the permutations of a string. E.g. permutation for ba, would be ba and ab, but what about longer st

Answer 1

Let's use input abc as an example.

Start off with just the last element (c) in a set (["c"]), then add the second last element (b) to its front, end and every possible positions in the middle, making it ["bc", "cb"] and then in the same manner it will add the next element from the back (a) to each string in the set making it:

"a" + "bc" = ["abc", "bac", "bca"]  and  "a" + "cb" = ["acb" ,"cab", "cba"] 

Thus entire permutation:

["abc", "bac", "bca","acb" ,"cab", "cba"]

Code:

public class Test 
{
    static Set permutations;
    static Set result = new HashSet();

    public static Set permutation(String string) {
        permutations = new HashSet();

        int n = string.length();
        for (int i = n - 1; i >= 0; i--) 
        {
            shuffle(string.charAt(i));
        }
        return permutations;
    }

    private static void shuffle(char c) {
        if (permutations.size() == 0) {
            permutations.add(String.valueOf(c));
        } else {
            Iterator it = permutations.iterator();
            for (int i = 0; i < permutations.size(); i++) {

                String temp1;
                for (; it.hasNext();) {
                    temp1 = it.next();
                    for (int k = 0; k < temp1.length() + 1; k += 1) {
                        StringBuilder sb = new StringBuilder(temp1);

                        sb.insert(k, c);

                        result.add(sb.toString());
                    }
                }
            }
            permutations = result;
            //'result' has to be refreshed so that in next run it doesn't contain stale values.
            result = new HashSet();
        }
    }

    public static void main(String[] args) {
        Set result = permutation("abc");

        System.out.println("\nThere are total of " + result.size() + " permutations:");
        Iterator it = result.iterator();
        while (it.hasNext()) {
            System.out.println(it.next());
        }
    }
}