I need to generate permutations over a set of integers.
However the placement of the numbers matter.
In position 0, the range is from 0-2. In position 1, the range is from 0-3. In position 2, the range is from 0-1.
So the answer would be:
000 001 010 011 020 021 030 031
100 101 110 111 120 121 130 131
200 201 210 211 220 221 230 231
Do something like:
public class Test {
public static void main(String[] args) {
Test x = new Test();
x.generatePermutations(2, 3, 1);
}
private void generatePermutations(int intervalOne, int intervalTwo,
int intervalThree) {
for (int i = 0; i <= intervalOne; i++) {
for (int j = 0; j <= intervalTwo; j++) {
for (int k = 0; k <= intervalThree; k++) {
System.out.print(i + "" + j + "" + k + " ");
}
}
}
}
}
Edit: I think this will do an array of any number of intervals: (You have to check its correctness though:
public class Test {
public static void main(String[] args) {
Test x = new Test();
int[] intervals = { 1, 1, 1, 1, 1 };
x.generatePermutations(intervals);
}
private void generatePermutations(int[] intervals) {
generatePermutations(intervals, 0, "");
}
private void generatePermutations(int[] intervals, int intpos,
String lastPerm) {
if (intpos == intervals.length)
return;
for (int i = 0; i <= intervals[intpos]; i++) {
if (intpos == intervals.length - 1) {
System.out.print(lastPerm + i + " ");
}
generatePermutations(intervals, intpos + 1, lastPerm + i);
}
}
}