ueg1990 - 5 months ago 9x

Java Question

I'm trying to find permutation of a given string, but I want to use iteration. The recursive solution I found online and I do understand it, but converting it to an iterative solution is really not working out. Below I have attached my code. I would really appreciate the help:

`public static void combString(String s) {`

char[] a = new char[s.length()];

//String temp = "";

for(int i = 0; i < s.length(); i++) {

a[i] = s.charAt(i);

}

for(int i = 0; i < s.length(); i++) {

String temp = "" + a[i];

for(int j = 0; j < s.length();j++) {

//int k = j;

if(i != j) {

System.out.println(j);

temp += s.substring(0,j) + s.substring(j+1,s.length());

}

}

System.out.println(temp);

}

}

Answer

Following up on my related question comment, here's a Java implementation that does what you want using the Counting QuickPerm Algorithm:

```
public static void combString(String s) {
// Print initial string, as only the alterations will be printed later
System.out.println(s);
char[] a = s.toCharArray();
int n = a.length;
int[] p = new int[n]; // Weight index control array initially all zeros. Of course, same size of the char array.
int i = 1; //Upper bound index. i.e: if string is "abc" then index i could be at "c"
while (i < n) {
if (p[i] < i) { //if the weight index is bigger or the same it means that we have already switched between these i,j (one iteration before).
int j = ((i % 2) == 0) ? 0 : p[i];//Lower bound index. i.e: if string is "abc" then j index will always be 0.
swap(a, i, j);
// Print current
System.out.println(join(a));
p[i]++; //Adding 1 to the specific weight that relates to the char array.
i = 1; //if i was 2 (for example), after the swap we now need to swap for i=1
}
else {
p[i] = 0;//Weight index will be zero because one iteration before, it was 1 (for example) to indicate that char array a[i] swapped.
i++;//i index will have the option to go forward in the char array for "longer swaps"
}
}
}
private static String join(char[] a) {
StringBuilder builder = new StringBuilder();
builder.append(a);
return builder.toString();
}
private static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
```

Source (Stackoverflow)

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