Hello.

How can I get all permutations in array[]. For example I have:

int[] array = {1, 2, 3};

As result I need:

123

132

213

231

312

321

Can you help me with that? Thanks in advance.

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- 04-07-2010, 11:24 AMmtKarray permutations
Hello.

How can I get all permutations in array[]. For example I have:

int[] array = {1, 2, 3};

As result I need:

123

132

213

231

312

321

Can you help me with that? Thanks in advance. - 04-07-2010, 11:25 AMr035198x
What have you tried?

- 04-07-2010, 11:30 AMmtK
This works perfectly with Strings, but I can't and can't get it work with an array. :/

Code:`public static void permutacija(String s) { permutacija("", s); }`

private static void permutacija(String prefix, String s) {

int N = s.length();

if (N == 0) {

System.out.println(prefix);

} else {

for (int i = 0; i < N; i++)

permutacija(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));

}

}

- 04-07-2010, 01:28 PMJosAH
There is a nice iterative algorithm that finds a next permutation given a current one; it finds permutations in lexicographic order (dictionary order), e.g. given the permutation 2, 3, 1 it finds the permutation 3, 1, 2. If the current permutation is 3, 2, 1 there is no next permutation; here's the algoritm:

1) find the largest i such that a[i] < a[i+1]

2) if no such i exists there is no next permutation.

3) find the largest j > i such that a[j] > a[i]; such j always exists

4) swap a[i] and a[j]

5) reverse the elements a[i+1], a[i+2] ... a[n] (all elements to the right of i).

I'm sure you can construct a bit of Java from this pseudo code.

kind regards,

Jos - 04-08-2010, 05:57 AMtoadaly
Assuming the size of the array is not fixed (if it is, you could just use nested loops), the best way is probably a re-entrant method.