Reputation: 1
I want to generate all possible teams of a group of n things taken k at a time, for example "abcd" = ab,ac,ad... without duplicates. I have written this, but it generates all permutations of a string. I have written a method to check if two strings have the same characters, but I don't know if this is the correct way.
package recursion;
import java.util.Arrays;
public class Permutations2 {
public static void main(String[] args) {
perm1("", "abcd");
System.out.println(sameChars("kostas","kstosa"));
}
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));
}
}
}
private static boolean sameChars(String firstStr, String secondStr) {
char[] first = firstStr.toCharArray();
char[] second = secondStr.toCharArray();
Arrays.sort(first);
Arrays.sort(second);
return Arrays.equals(first, second);
}
}
Upvotes: 0
Views: 113
Reputation: 909
This should work without recursion:
private static void perm(String s) {
char[] arr = s.toCharArray();
for (int i = 0; i < s.length() - 1; i++) {
for(int j = i + 1; j < s.length(); j++) {
System.out.println(String.valueOf(arr[i]) + String.valueOf(arr[j]));
}
}
}
It's O(n**2).
Upvotes: 0
Reputation: 42040
May you want the Enumerating k-combinations.
private static void combinate(String s, String prefix, int k) {
if (s.length() < k) {
return;
} else if (k == 0) {
System.out.println(prefix);
} else {
combinate(s.substring(1), prefix + s.charAt(0), k - 1);
combinate(s.substring(1), prefix, k);
}
}
public static void main(String[] args) {
combinate("abcd", "", 2);
}
Output:
ab
ac
ad
bc
bd
cd
See Introduction to Programming in Java - Recursion - Robert Sedgewick and Kevin Wayne
Upvotes: 0
Reputation: 53839
You are describing a power set.
Guava has an implementation.
Upvotes: 3