问题
Having problems trying to show every combination of a character of array without repeating letters.
public static String[] getAllLists(String[] elements, int lengthOfList)
{
//initialize our returned list with the number of elements calculated above
String[] allLists = new String[(int)Math.pow(elements.length, lengthOfList)];
//lists of length 1 are just the original elements
if(lengthOfList == 1) return elements;
else
{
//the recursion--get all lists of length 3, length 2, all the way up to 1
String[] allSublists = getAllLists(elements, lengthOfList - 1);
//append the sublists to each element
int arrayIndex = 0;
for(int i = 0; i < elements.length; i++)
{
for(int j = 0; j < allSublists.length; j++)
{
//add the newly appended combination to the list
allLists[arrayIndex] = elements[i] + allSublists[j];
arrayIndex++;
}
}
return allLists;
}
}
The above code works perfect but use's each letter more than once which cant be done in this case.
And i am stuck how to do this now.
回答1:
Here's an example implementation. Essentially it takes a String and iterates over every character, putting that character at the front. It then recurses on the remaining characters. That structure removes your issue of repeated letters, because the input to the recursion has removed the character you've already used.
I also stored results in a set to remove semantic equivalences. The input 'aab' can switch char 0 and char 1 but still be 'aab.' I used a TreeSet to preserve ordering for easier verification of the output, but HashSet would be faster.
public static Set<String> permute(String chars)
{
// Use sets to eliminate semantic duplicates (aab is still aab even if you switch the two 'a's)
// Switch to HashSet for better performance
Set<String> set = new TreeSet<String>();
// Termination condition: only 1 permutation for a string of length 1
if (chars.length() == 1)
{
set.add(chars);
}
else
{
// Give each character a chance to be the first in the permuted string
for (int i=0; i<chars.length(); i++)
{
// Remove the character at index i from the string
String pre = chars.substring(0, i);
String post = chars.substring(i+1);
String remaining = pre+post;
// Recurse to find all the permutations of the remaining chars
for (String permutation : permute(remaining))
{
// Concatenate the first character with the permutations of the remaining chars
set.add(chars.charAt(i) + permutation);
}
}
}
return set;
}
Example run:
public static void main(String[] args)
{
for (String s : CharPermuter.permute("abca"))
{
System.out.println(s);
}
}
Generates:
aabc
aacb
abac
abca
acab
acba
baac
baca
bcaa
caab
caba
cbaa
来源:https://stackoverflow.com/questions/9666903/every-combination-of-character-array