/
*************************************************************************
* Compilation: javac Permutations.java
* Execution: java Permutations N
*
* Enumerates all permutations on N elements.
* Two different approaches are included.
*
* % java Permutations 3
* abc
* acb
* bac
* bca
* cab
* cba
*
*************************************************************************/
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) {
System.out.println(a);
return;
}
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);
perm1(elements);
System.out.println();
perm2(elements);
}
}
On May 8, 9:54 am, Shoubhik <[email protected]> wrote:
> Could somebody post an algorithm/code for obtaining permutations of an
> n-letter string .
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